Diamonds in mathematical inequalities

Diamonds in mathematical inequalities 1

Proposal for a book

Tran Phuong

Director of Center for Research and Support Development of Intellectual Products.

No 19/89, Thai Ha Street, Dong Da District, Hanoi, Viet nam.

Email: [email protected]; [email protected]

Phone No:

(844) 5-376-340; (84)913-220-457

Assistants’ mobile phone numbers:

0061402201501

Email:[email protected]

Fax:

(84)(04) 943-5816

(84)(04)719-9641


��

.

I.1 Title

Diamonds in mathematical inequalities.

I.2. Purpose of the book The book aims, first, to provide students with a comprehensive and minute system of

typical inequality demonstration methods and techniques, ranging from classical to modern

ones, which, due to the fact that their importance remains unchanged throughout the flow of

time, can be considered as “diamonds in mathematical inequalities”. This main purpose is

accomplished by a number of valuable mathematic samples with interesting “lead-ins” and,

most importantly, a dialectic viewpoint on each solving method.

The major goal of the authors is, therefore, to helping students to acquire multitude of

mathematical tools and methods in solving inequalities, then, be able to achieve excellence in

this field. Besides, on utilizing a new approach in presenting familiar mathematical facts, the

authors also hope to highlight in readers’ mind the importance of a good mathematical

inequality-based thinking in developing their creativeness as well as their ability to critically

evaluate changes in life.

The success of this book will also mean the publication of the authors’ other books on

other fields of secondary mathematics.

I.3. Why is this book different from others?

• ��

“ Money ? Eventually spent

Beauty ? Eventually faded...

Only Intellect rooted in mind

And feelings rooted in heart Will last

forever with time

(Tran Phuong, 1990)


Comparisons in life The material and spiritual world in the universe is always on the move, ever changing to

create different things and phenomena. This diversity can be seen through opposite images:

big − small ; wide − narrow ; long − short ; high − low ; plentiful − scarce ; rich − poor ;

beautiful − ugly and so on. Extensively, it is that the divergence between Buddhism,

Mohammedanism and Catholicism; the differences between the Oriental − Occidental

cultures and between the major philosophic thoughts all over the world.

Everybody′s life is an ongoing search to set his/her own values. Every thing gets its

standing in the ever changing world by its own values; however, one does not often realize that everything can only obtain its value in comparison with some other thing. It is this relationship that creates inequalities.

In fact, though "all comparisons are odious" as the saying goes, men cannot help resorting to comparison to evaluate things. A kilogram of Korean ginseng is 1000 times smaller than a ton of rice in term of mass, but costs 10 times more. Take the example of a president or a car driver, who is more important? You may say that the head of state is certainly more important than a racer, but only in state affairs. In respect of annual income, former President Clinton's US $200,000 annual salary in office is derisory in comparison with world famous formula 1 car racer Michael Schumacher's US $60,000,000 annual income. Thus, factors in a comparison should be identified, quantified and converted to a same kind of unit - whether you want to or not.

In order to develop comparison thinking and evaluation in life, notions of mathematical inequalities are given us as early as school age. Children at kindergarten learn to compare 1 with 2 to have a 1 < 2 inequality. Fourth formers begin to achieve more difficult comparisons,

for instance 135143

with ��

��, etc. Some of them proceed by reduction to a common

denominator, which is rather complicated, i.e.

��

��

× ×= = = =× ×

but ��

�� < �

��

�� <

Others intelligently operate by fractions complementation as follows:

��

�� +

� � ��

�� = = + ; now

� �

�� > hence

��

�� <

Comparison problems grow more and more difficult with more extensive operations. All mathematicians share the common concept "Basic results of mathematics often are expressed by inequalities in stead of equalities". The same is true in real life, where one always encounters differences between things and phenomena, and even changes in a single phenomenon by the minute. Indeed, if we are subject to no change after every second, then, according to the induction principle, 50 years later, we would not get any older. On the other


hand, in social life, inequality-based thinking is always needed to assess business activities, export-import industry, stock exchange market, finances, banking... Therefore, in order to develop thinking and properly assess the changes in life, a good mathematical inequality-based thinking is most necessary. With these introductory words, the author wishes that the readers consider the subject treated in this book as a close and intimate one.”

With such profound philosophical understandings, the writer has always been aware of the necessity to ensure an inspiration and intellectual delight in each technique and method presented. The book, there fore, will encourage students to acquire a multitudes of inequality demonstration methods as effective tools not only in solving mathematics problems but also in critically evaluate changes in life.

• ��

1. The classical “diamonds” were introduced in new angles of view with typical

techniques and methods followed by a variety of interesting examples ranging from simple to

complex.

To be more specific, the authors insufflate in each inequality demonstration technique a

new notion imbued with a forceful philosophic connotation, the notion of “point of incidence”. The utility of such creative penmanship is rooted from the ideas relating to the

point of incidence in the film “Teheran 43”. The film, which was produced by the cooperation

of Russia, France and Italia, has the setting of the summer 1943, when the World War II was

growing fiercer and fiercer. In the context, the U.S., Great Britain and the Soviet Union

envisaged opening a Front of Alliance and the Germans plotted to assassinate the three leaders

of the Alliances, Roosevelt, Churchill and Stalin. The German spy Schiller trusted professional

hit-man Max with the assassination, but bade him to work out the venue and the time of the

meeting. From the military developments, Max estimated that the meeting would take place at

the end of 1943. Added with diplomatic pretext for the meeting, Max came to the final

conclusion that it must take place no sooner or later than Churchill’s birthday, which was 30th,

November and the venue must be in a British colony far away from the front in Europe. The

venue, accordingly, must be no other than the British Embassy in Teheran, the Iranian

Capital. From such estimation Max flew to Teheran three months before the meeting and

hired locals to dig a tunnel ending up in the British Embassy.

Thus, the idea of choosing points of incidence in life springs from considering the

evolution and development of things to estimate where they would lead to, with a view to

orienting the direction of approach from the outset. As to the techniques of choosing the point

of incident in equalities, it is based on the state of variables when the two terms of the

inequalities happen to equal. Typical examples illustrating this technique are found precisely

presented in the very first introduction on AM-GM, Cauchy-Schwarz, Bernoulli inequalities

and throughout this book.


2. The modern “diamonds”, of which the typical ones are ABC, GTA, GLA, DAC

methods, are discussed in details so that readers can easily grasp the spirit of each technique

and method and apply them.

After all, this is a profound book on the methods and techniques of mathematical

inequality demonstration. The problems given are arranged didactically upwards, i.e.

gradually from lower to higher levels. A variety of typical examples are given to illustrate

each problem; exercises for readers also range from simple to complex ones with large

amplitude. All of them are presented in an interesting way but easy to understand and apply.

Hopefully, readers would feel easy to understand and to perceive this book, and to

evaluate the authors’ “mathematical creations”. It convey new perspectives and view points

to many familiar problems and challenges its audience with a number of new ones. In one

word, with this book, readers may have chances to develop and to make themselves

recognized.

3. Mathematics is a language itself and also Inequality should be a public language that

can connect everybody throughout the world easily. Therefore the authors used many

Mathematics notations to reduce the English language so that even those who don’t know

English well can also enjoy this book.

I.4. Table of contents

The book consists of 5 chapters, for a total of about 480 pages. The major content of the book

can be summarized as follows:

Chapter I: Diamonds in classical mathematical inequalities

§1. AM-GM inequalities

§2. Cauchy-Schwarz inequalities

§3. Holder inequalities

§4. Minkowski inequalities

§5. Chebyshev inequalities

Chapter II: Diamonds in modern mathematical inequalities

§6. Schur inequalities

§7. Muirhead inequalities

§8. Permutation inequalities

Chapter III: Diamonds of analytic method

§9. Fermat theorem (Derivative method)

§10. Lagrange theorem

§11. Bernoulli inequalities


§12. Jensen inequalities

§13. Karamata inequalities

§14. Vasile Cirtoaje inequalities (RCF, LCF and LCRCF theorem)

§15. Popoviciu inequalities

§16. Riman theorem (Integral method)

Chapter IV: Diamonds in contemporary inequalities

§17. UCT method

§18. SOS method

§19. GMV method

§20. ABC method

§21. Equal variable method

§22. Geometricalize Algebra method

§23. Divide and conquer method

Chapter V: Some creations on mathematical inequalities

§24. Selective papers on inequalities

§25. Nice solutions to selective inequalities

§26. Challenging problems

I.5. Who is the audience?

The book is intended for a wide range of audience. It is first and particularly meant for

capable basic secondary students, candidates to national and international mathematics

contests, mathematics teachers at all levels and researchers. On completing this book, the

authors, however, did not target the public from only one country. Believing that

mathematics language does not change across national boundaries, they hope that the book

will be able to reach an international readership and prove to be helpful for everyone who is

seeking ways to broaden their mathematical horizons.

I.6. Final comment

This book is the authors’ soul-felt work achieved through years of hard work, based on

their vast experience in mathematics and mathematical education. It is a comprehensive work

in a field favorable to observation, intuition and creativeness, all of which are outstanding

talents of the Vietnamese people. Having devoted themselves entirely to completing this book,

the authors are convinced that it will last as long as a close friend of the world’s

mathematics lovers.


II. ��The initial section of Sample Work is a part of chapter one of the book, which was

written with a brand new style for a well – known method: “Point of incidence” In this

chapter, we will introduce the technique of choosing the “point of incidence” in AM – GM,

Cauchy-Schwarz and Bernoulli inequalities. However, it is noted that selecting “point of

incidence” is only the typical technique among a total of 30 techniques of using AM – GM

inequalities presented in the manuscript of the book “Collections of Topics, Methods and

Techniques in Algebraic Inequality Demonstration”, which is 2222-pages thick and was

completed by Tran Phuong due to the impetus raised in his mind on the occasion of President

W.J. Clinton visited Vietnam on November, 16th, 2000. In this part, the ideas relating to the

point of incidence in the film “Teheran 43” will be mentioned again.

In the remaining section, we will consider the best estimation of Nesbit −−−− Shapiro

Inequality in primary mathematics.

In respect of format, we would like to note here that in the hope of presenting the content

of the book in the most convenient way, we choose the paper size of 17×24cm and the page

layout of 14×21cm, which is very common in Vietnam.

INTRODUCTION: FROM THE STORY ABOUT “POINTS OF INCIDENCE” TO POINT OF INCIDENCE IN MATHEMATICAL INEQUALITY

IDEAS RELATING TO “POINTS OF INCIDENCE” IN TEHERAN 43

The film, which was produced by the cooperation of Russia, France and Italia, has the setting

of the summer 1943, when the World War II was growing fiercer and fiercer. In the context, the

U.S., Great Britain and the Soviet Union envisaged opening a Front of Alliance and the Germans

plotted to assassinate the three leaders of the Alliances, Roosevelt, Churchill and Stalin. The

German spy Schiller trusted professional hit-man Max with the assassination, but bade him to

work out the venue and the time of the meeting. From the military developments, Max

estimated that the meeting would take place at the end of 1943. Added with diplomatic

pretext for the meeting, Max came to the final conclusion that it must take place no sooner or

later than Churchill’s birthday, which was 30th, November and the venue must be in a British

colony far away from the front in Europe. The venue, accordingly, must be no other than the

British Embassy in Teheran, the Iranian Capital. From such estimation Max flew to Teheran

three months before the meeting and hired locals to dig a tunnel ending up in the British Embassy.


Thus, the idea of choosing points of incidence in life springs from considering the

evolution and development of things to estimate where they would lead to, with a view to

orienting the direction of approach from the outset. As to the techniques of choosing the point

of incident in equalities, it is based on the state of variables when the two terms of the

inequalities happen to equal. Typical examples illustrating this technique will be precisely

presented in the very first introduction on AM-GM, Cauchy-Schwarz, Bernoulli inequalities

and throughout this book.

POINT OF INCIDENCE IN EQUALITIES

Setting the manner:

In proving the inequality A ≥≥≥≥ B we often follow either the two following patterns:

Pattern 1: Create a sequence of intermediary inequalities

A ≥≥≥≥ A1 ≥≥≥≥ A2 ≥≥≥≥... ≥≥≥≥ An −−−− 1 ≥≥≥≥ An ≥≥≥≥ B

Pattern 2: Create a sequence of component inequalities

1 1

2 2

. . . . . . . .

n n

A B

A B

A B

A B

≥�� ≥�+�� ≥�

� ≥

or

1 1

2 2

0

0

0

. . . . . . . . . . . .

n n

A B

A B

A B

A B

≥ ≥�� ≥ ≥�×�� ≥ ≥�

� ≥

To create intermediary or component inequalities we need to note that if the 'Central

inequality A ≥≥≥≥ B' becomes 'A = B' at a standard P , all the intermediary inequalities in

Pattern 1 or all the component inequalities (local inequality) in Pattern 2 also become

equalities. To find standard P we need to pay attention to the symmetry of variables and the

conditions for equality to occur in AM − GM Inequality where all joining variables are

equal. For estimating at what standard 'A = B' occurs is to guide algebraic transformations and

estimations of intermediary or component inequalities, the work can be called 'Inspecting the

condition of equality occurring and point of incidence in the inequality'.

We will be more familiar with this idea in a variety of examples ranging from single to

complex in the detailed discussion of sample exercises later.


III. ��

III.1. Tran Phuong, the chief author of the book, is the Director of Center for Research and

Development Support of Intellectual Products in Vietnam.

Like most authors of mathematics books, Tran Phuong received systematic and

professional education and training specializing in this field. Before entering university, he

studied in Mathematics Specialized School at the University of Natural Sciences, Hanoi, the

cradle in which up to 60% of Vietnamese IMO medalists were trained and nourished so far.

After his school years, he continued pursuing his interest in mathematics in Teacher Training

University, Hanoi, the leading center for training mathematics professors in Vietnam.

After graduation, the author, however, pursued his career in his own way which is

teaching on the invitations of high schools, universities and mathematics centers in Vietnam.

During the years from 1990 to 2000, up to approximately 10 thousands of students attended

his classes on mathematics. The majority of them were excellent pupils, including many

candidates to both national and international Mathematics contests. Vietnamese IMO 43 gold

medalist Pham Gia Vinh Anh is among his best pupils. Since his ultimate goal in teaching

mathematics is student’s full development in mathematics-based thinking, not merely the

mathematics knowledge itself, many of his students, on their outcome of what they acquired

in his classes, contribute actively in various fields of the Vietnamese society.

On realizing his dream of devoting entirely to mathematics science teaching and learning,

Tran Phuong has had a great passion for writing books on mathematics, especially on

inequalities. His first book, which was published in 1993, was a comprehensive work on

mathematical inequalities. From then on, he has completed a total of 10 books, all of which

were introduced by five first-ranked publishers in Vietnam (Education Publisher, Youth

Publisher, Knowledge Publisher, Ho Chi Minh Publisher and Da Nang Publisher) and soon

became the best-sellers among referenced books for secondary pupils in Vietnam. He is also

the author of an intellectual game show “Vietnamese Infant Prodigy” on television.

For his great contribution to mathematics teaching and learning in Vietnam, he was

awarded the noble “Vietnam’s Genius 2006” by Vietnamese government.


III.2. Contributors:

1. Tran Tuan Anh

PhD Candidate, Shool of Mathemmatics Georgia Institute of Technology,

Atlanta,GA 30332, USA

2. Pham Gia Vinh Anh

Bachelor of Information Technology, University of Sydney, Australia,

Gold Medal in IMO 43

3. Nguyen Anh Cuong

Student of School of Computing, Singapore National University, course 2006-2010

4. Bui Viet Anh

Student of Department of Electronics and Telecommunications , University of

Technoly, Ha Noi,course 2005-2009, Vietnam

5. Le Trung Kien

Mathematics majored students, school period of 2004 - 2007, the Gifted High

School of Bac Giang, Vietnam

6. Phan Thanh Viet


Shool of Luong Van Chanh, Phu yen, Vietnam

7. Vo Quoc Ba Can


School of Ly Tu Trong, Can Tho, Vietnam

8. Hoang Trong Hien


School of Le Hong Phong, Nam Dinh, Vietnam

9. Nguyen Thuc Vu Hoang


School of Le Quy �on, Quang Tri,Vietnam

10. Nguyen Quoc Hung

Academic Period of 2005 -2009, Faculty of Mathematics & Informatics,

University of Natural Sciences, Ho Chi Minh City, Vietnam

Diamonds in mathematical inequalities

11

CHAPTER ONE: DIAMONDS IN

CLASSICAL MATHEMATICAL INEQUALITIES

§1.1. DIFFERENT COLORS IN CATCHING POINT OF INCIDENT

TECHNIQUE IN AM – GM INEQUALITY

Main points:

I. AM −−−− GM inequality

1. General form

2. Special cases

3. Proof

II. Different colors in catching point of incident technique in AM – GM inequality

1. Point of incidence in evaluating from AM to GM

2. Point of incidence in evaluating from GM to AM

3. Point of free incidence or homogeneous principle in AM – GM inequality

4. Specialization in inequality of same degree 4. Specialize in homogeneous inequality

5. Non-symmetric point of incidence in AM – GM inequality

6. Balancing Coefficient Technique (Method of equalizing coefficient )

7. Using AM – GM in homogeneous inequality Applying AM – GM to inequality of different degree

8. Specialize in un-homogeneous inequality Specialization in inequality of different degree

9. The most beautiful solutions for four trigonometry inequalities

10. Selective problems in using point of incidence for AM – GM

I. AM −−−− GM INEQUALITY

�� Suppose 1 2, , ... na a a are n non-negative real numbers, then:

1.1. Form 1: 1 21 2

......n n

nn

a a aa a a≥

+ + +

1.2. Form 2: 1 2 1 2.... ...nn nna a a a a a≥+ + +

1.3. Form 3: 1 21 2

.......

n

nn

a a aa a a

n

+ + +� �≥� �

� �

1.4. Equality occurs ⇔ 1 2 ... 0na a a= = = ≥

1.5. Corollary

• If 1 2, , ..., 0na a a ≥ and 1 2 ... na a a S+ + + = is constant, then

( )1 2Max ...

n

n

Sa a a

n

� �= � ��

occurs 1 2 ... n

Sa a a

n= = = =⇔

• If 1 2, , ..., 0na a a ≥ and 1 2 ... na a a P= is constant, then

( )1 2Min ... .nna a a n P+ + = occurs 1 2 ... n

na a a P⇔ = = = =

Point of incident technique in AM-GM inequality

12

��

n

Form n = 2 n = 3 n = 4

Condition ∀ a, b ≥ 0 ∀ a, b, c ≥ 0 ∀ a, b, c, d ≥ 0

Form 1 2

a bab

+≥ 3

3

a b cabc

+ +≥ 4

4

a b c dabcd

+ + +≥

Form 2 2.a b ab+ ≥ 33.a b c abc+ + ≥ 44.a b c d abcd+ + + ≥

Form 3

2

2

a bab

+� �≥� �

� �

3

3

a b cabc

+ +� �≥� �

� �

4

4

a b c dabcd

+ + +� �≥� �

� �

Equality a = b a = b = c a = b = c = d

�� 1 21 2

......n n

nn

a a aa a a≥

+ + + ,

1 2, , ... 0

na a a∀ ≥ (1)

There are 36 solutions for this inequality; following is the proof by mathematic induction

• For n = 2: ( )

2

1 21 2 1 2

1 2 1 202 2 2

a aa a a aa a a a

−+ +− = ≥ � ≥

Equality occurs ⇔ a1 = a2. Suppose the inequality is true for n ≥ 2.

• We will prove the inequality is true for (n + 1) numbers 1 2 1, , ... , 0n na a a a + ≥

Using the inductive hypothesis for n numbers 1 2, , ... 0na a a ≥ we have

( ) 1 2 11 2 1

1

. ......

1 1

nn nn n

nSn a a a aa a a a

n n

++

+ = ≥++ + + +

+ +. Let

( 1)1 2

11

...

for , 0

n n

n

nn

a a a p

a q p q

+

++

� =

= ≥�

� 1 1

11

n n

n

np qS

n

+ +

+

+≥

+ (1). We will prove

1 11

1 2 1... (2)1

n nn n

n n

np qp q a a a a

n

+ ++

+

+≥ =

+

We have: ( )1 1 1

( )1 1

n nn n n nnp q

p q np p q q p qn n

+ ++ � − = − − −� �+ +

( ) ( )2

1 2 2 3 2 1 2 1( )( ) ... . ... ... 0

1

n n n n n n n np qp p p q p p p q q p p q q

n

− − − − − − − −− � = + + + + + + + + + + ≥� �+

From (1) and (2) � 1 2 1 11 2 1

......

1

n n nn n

a a a aa a a a

n

+ ++

+ + + +≥

+

Equality occurs 1 2

1 2 1

......

n

n n

a a aa a a a

p q+

= = =�⇔ ⇔ = = = =

=�

According to the induction principle, the inequality is true for ∀n ≥ 2, n ∈ �


13

II. CATCH POINT OF INCIDENCE TECHNIQUE IN AM – GM INEQUALITY

1. POINT OF INCIDENCE IN EVALUATING FROM AM TO GM

Leading in:

From a simple problem: Find the minimum value of S = a b

b a+ for a, b > 0

We can find out the solution: 2 2a b a b

Sb a b a

= + ≥ ⋅ = . When a = b > 0, Min S = 2 but when we

consider the problem in a different domain we will have an interesting problem connecting

with fix point of incidence in AM – GM inequality.

Problem 1. Given a ≥ 3. Find the minimum value of S = 1aa

+

Solution

• Common mistake: 1 12 2 Min 2S a a Sa a

= + ≥ ⋅ = � =

• Cause: 1Min 2 1S aa

= ⇔ = = contradicts the assumption a ≥ 3

• Analyzing and finding the solution:

Consider the variation table of 1,aa

and S to estimate Min S

a 3 4 5 6 7 8 9 10 11 12 �� 30

1a

13

14

15

16

17

18

19

110

111

1

12 ��

130

S 133

144

155

166

177

188

199

11010

11111

11212

�� 13030

As we can see from the Variation Table, if a increases, S increases and then it can be

estimated that S receives the least value at a = 3. We say = ��

�Min occurs at the point of

incidence a = 3.

Since in AM −−−− GM Inequality the equality occurs on the condition that all joining numbers

are equal, at the point of incidence a = 3 we cannot use AM −−−− GM Inequality directly for two

numbers a and 1a

because 3 ≠ 13

. We assume that AM −−−− GM Inequality is used for the

couple ( )1,a

aα such that at the point of incidence a = 3 occur 1a

a=

α that means the following

‘Point of incidence’ Pattern holds:

Pattern:

3

1 33 9

31 13

a

a

a

� =α α

= � � = � α =α =

�

: point of incidence coefficient

We transform S according to the 'Point of incidence' Pattern as above.


14

• Right solution: ( ) 8 8.3 101 1 12.9 99 9 3

aa aS aa a a

= + = + + ≥ ⋅ + = . For a = 3 then 10Min3

S =

Problem 2. Given 0

1

a,b

a b

>�

+ ≤� Find the minimum value of the expression 1ab

ab= +S

Solution

• Common mistake : 1 12. 2 Min 2S ab ab Sab ab

= + ≥ ⋅ = � =

• Cause: 1 1 1Min 2 1 1 12 2 2

a bS ab abab

+= ⇔ = = � = ≤ ≤ � ≤ : illogical


The expression S contains two variables a, b but if we take t ab= or 1tab

= , the expression

1S tt

= + will contain only one variable. When changing the variables we must find the

defined area of the new variables as follows:

1 1t ab

ab t= � = and

( ) ( )2 2

1 1 14

122

tab a b

= ≥ ≥ =+

Problem will be: Given t ≥ 4. Find the minimum value of the expression 1S tt

= +

� 'Point of incidence' Pattern:

4t = �

4

1 1

4

t

t

�=α α

=�

� 1 4

4=

α � 16α = : Point of incidence coefficient

� General solution:

1 15 151 12.

16 16 16 16

t t t tS t

t tt

� �= + = + + ≥ ⋅ +� �

� �

2 15 2 15.4 17

4 16 4 16 4

t= + ≥ + =

For t = 4 or 1

2a b= = then

17Min

4S =

� Reduced solution: Since t = 4 ⇔ 1

2a b= = we transform S directly as follows

1 1 15

16 16S ab ab

ab ab ab

� �= + = + +� �

� � ≥

( )2

1 15 172.

16 416

2

abab a b

⋅ + ≥+

For 12

a b= = then 17Min4

S =


15

Problem 3. Given

0

3

2

a,b,c

a b c

>�

+ + ≤�

Find the minimum value of 1 1 1

a b ca b c

= + + + + +�

Solution

• Common mistake: 61 1 1 1 1 16. 6 Min 6S a b c abc S

a b ca b c= + + + + + ≥ ⋅ ⋅ ⋅ = � =

• Cause:

1 1 1Min 6 1S a b ca b c

= ⇔ = = = = = = � 3

32

a b c+ + = > contradicts the assumption.


Since S is the symmetrical expression of a, b, c we estimate Min S occurs at 12

a b c= = =

� 'Point of incidence':

1

2a b c= = = �

1

2 1 2

21 1 1 2

a b c

a b c

�= = =

� =

α= = =α α α α�

� 4α = : Point of incidence coefficient

� Method 1: 1 1 1 1 1 1 3 1 1 1

4 4 4 4S a b c a b c

a b c a b c a b c

� � � �= + + + + + = + + + + + + + +� � � �

� � � �

363

1 1 1 3 1 1 1 9 16. 3. 3

4 4 4 4 4abc

a b c a b c abc

� �≥ ⋅ ⋅ ⋅ + ⋅ ⋅ = + ⋅� ��

� �

9 1 27 1 27 1 15

3 3 334 4 4 2

3 2a b c a b c

≥ + ⋅ = + ⋅ ≥ + ⋅ =+ + + +

. For 1

2a b c= = = then

15Min

2S =

Problem 4. Given

0

32

a,b,c

a b c

>�

+ + ≤�

Find Min of 2 2 2

2 2 2

1 1 1a b c

b c a= + + + + +�

Solution

• Common mistake: 2 2 2 2 2 23 62 2 2 2 2 2

1 1 1 1 1 13. 3.S a b c a b c

b c a b c a

� � � � � �≥ + ⋅ + ⋅ + = + + +� � � � � �

� � � � � �

62 2 26

2 2 2

1 1 13. 2 2 2 3. 8 3 2 Min 3 2a b c S

b c a

� � � � � �≥ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = = � =� � � � � ��

� � � � � �

• Cause: 31 1 1Min 3 2 1 32

S a b c a b ca b c

= ⇔ = = = = = = � + + = > contradicts the assumption


Since S is the symmetrical expression of a, b, c we estimate Min S occurs at 12

a b c= = =


16


1

2a b c= = = �

2 2 2

2 2 2

1

4

1 1 1 4

a b c

a b c

�= = =

= = = αα α α�

� 1 4

4=

α � 16α = : Point of incidence coefficient

2 2 2

2 2 2 2 2 2

16 terms 16 terms 16 terms

1 1 1 1 1 1

16 16 16 16 16 16S a b c

b b c c a a= + + ⋅ ⋅ ⋅ + + + + ⋅ ⋅ ⋅ + + + + ⋅ ⋅ ⋅ +

��

2 2 2

17 17 17

16 32 16 32 16 3217. 17. 17.

16 16 16

a b c

b c a≥ + + 17 17 17

8 16 8 16 8 1617

16 16 16

a b c

b c a

� �= + +� ��

� �

3 17 17 17 178 16 8 16 8 16 8 5 5 5

117 3 3 17

16 16 16 16

a b c

b c a a b c

� � �≥ ⋅ ⋅ ⋅ = ⋅� ��

( )17 5 15

17

3 17 3 17 3 17

22 (2 .2 .2 ) 2 2 223

a b c a b c

= ≥ ≥⋅ + +⋅

. For 1

2a b c= = = then

3 17Min

2S =

Problem 6. Let 2 2 2

, , 0

1

a b c

a b c

≥�

+ + =�

. Find the minimum value of T = 1a b cabc

+ + +

(Macedonia 1999)

Solution

• Common mistake: 41 14 4a b c a b cabc abc

+ + + ≥ ⋅ ⋅ ⋅ ⋅ = � Min T = 4

• Cause: Min T = 4 ⇔ 1 1a b c a b cabc

= = = ⇔ = = = contradicts the assumption.


Estimate that the point of incidence of Min T is 1

3a b c= = = , then 1 3 3

abc=


1

3a b c= = = �

1

3 3 3 1

31 3 3

a b c

abc

� = = =

� =α

= αα�

� 9α = : Point of incidence coefficient

• Right solution: 43

2 2 2

8 8 81 1 44 4 39 9 9 3 3

93

a b c a b cabc abc abc

a b c

+ + + + ≥ ⋅ ⋅ ⋅ ⋅ + = + =� �+ +� ��


17

2. POINT OF INCIDENCE IN EVALUATING FROM GM TO AM

Remark: Consider AM – GM inequality: 1 2

1 2

terms

......n

nn

nn

a a aa a a

+ + +≥��

, 1 2, , ... 0

na a a∀ ≥

Notices that in RHS expression GM we can see that the root index and the numbers of factors

inside the root are equal (both equal to n). Therefore, if these two quantities are different from

each other, they need adjusting towards being equal. In general, we usually deal with the problems

that we need to multiply the expressions with suitable constant so that the numbers of factors

inside the root are equal to the root index and suitable with the speculated point of incidence.

Problem 1. Given 0

1

a,b,c

a b c

≥�

+ + =� Find the minimum value of S = 3 3 3a b b c c a+ + + + +

Solution

• Common mistake: ( )

( )

( )

3 3

3 3

3 3

1 1.1.13

1 1.1.13

1 1.1.13

a ba b a b

b cb c b c

c ac a c a

+ + +� + = + ≤ + + ++ + = + ≤

+ + + + = + ≤�

3 3 3 2( ) 6 8

3 3

a b cS a b b c c a

+ + +� = + + + + + ≤ =

8Max

3S� =

• Causes of the mistake:

18

13

1

a b

Max S b c

c a

+ =�

= ⇔ + + = + =�

2( ) 3 2 3a b c� + + = ⇔ = � illogical

• Estimation of the Point of incidence of Max S:

Since S is the symmetrical expression for a, b, c Max S often occurs when

1

a b c

a b c

= =�

+ + =� ⇔

1 233

a b c a b b c c a= = = ⇔ + = + = + =

• Right solution:

3 3 33

3 3 33

3 3 33

2 2( )

9 2 2 9 3 3( )4 3 3 4 3

2 2( )

9 2 2 9 3 3( )4 3 3 4 3

2 2( )

9 2 2 9 3 3( )4 3 3 4 3

a b

a b a b

b c

b c b c

c a

c a c a

�+ + +

+ = ⋅ + ⋅ ⋅ ≤ ⋅

+ + ++ + = ⋅ + ⋅ ⋅ ≤ ⋅ + + +

+ = ⋅ + ⋅ ⋅ ≤ ⋅�

3 3 3 33 32( ) 49 9 6

184 3 4 3

a b cS a b b c c a

+ + +� = + + + + + ≤ ⋅ = ⋅ =

For 2 13 3

a b b c c a a b c+ = + = + = ⇔ = = = , 3Max 18S =


18

Problem 2. Given 0

3

a , b , c

a b c

>�

+ + =� Find the greatest value of the expression:

( ) ( ) ( )3 3 32 2 2a b c b c a c a b= + + + + +�

Solution



13

a b ca b c

a b c

= =�⇔ = = =

+ + =� �

3 3 3 3

2 2 2 3

a b c

b c c a a b

= = =�

+ = + = + =�

• Right solution:

( ) ( )( )

( ) ( )( )

( ) ( )( )

3 3

3 3

3 3

3 3

3 3

3 3

3 2 31 12 3 2 .339 9

3 2 31 12 3 2 .339 9

3 2 31 12 3 2 .339 9

a b ca b c a b c

b c ab c a b c a

c a bc a b c a b

� + + ++ = ⋅ + ≤ ⋅

+ + +

+ = ⋅ + ≤ ⋅+ + + +

+ = ⋅ + ≤ ⋅�

� ( ) ( ) ( ) 33 3 3

3 3

6( ) 91 12 2 2 9 3 339 9

a b cS a b c b c a c a b

+ + += + + + + + ≤ ⋅ = ⋅ = ⋅

For 1a b c= = = , 3Max 3. 3S =

Problem 3. Given 2 2 2

0

12

a , b , c

a b c

>�

+ + =�

Find the greatest value of the expression:

3 3 32 2 2 2 2 2a b c b c a c a b= ⋅ + + ⋅ + + ⋅ +�

Solution



2 2 2

02

12

a b ca b c

a b c

= = >�⇔ = = =

+ + =�

�

2 2 2

2 2 2 2 2 2

2 2 2 8

8

a b c

b c c a a b

� = = = + = + = + =�

• Right solution:

( ) ( ) ( )( )

( ) ( ) ( )( )

( ) ( ) ( )( )

2 2 22 3 26 63 2 2 6 2 2 2 2 2

2 2 22 3 26 63 2 2 6 2 2 2 2 2

2 2 22 3 26 63 2 2 6 2 2 2 2 2

6 2 81 12 .82 2 6

6 2 81 12 .82 2 6

6 2 81 12 .82 2 6

a b ca b c a b c a b c

b c ab c a b c a b c a

c a bc a b c a b c a b

� + + +⋅ + = + = ⋅ + ≤ ⋅

+ + ++ ⋅ + = + = ⋅ + ≤ ⋅ + + + ⋅ + = + = ⋅ + ≤ ⋅�

� ( )2 2 2

3 3 32 2 2 2 2 2 10 241 122 6

a b cS a b c b c a c a b

+ + += ⋅ + + ⋅ + + ⋅ + ≤ ⋅ =

For 2a b c= = = , Max 12S =


19

Problem 4. Let be given a ≥ 2; b ≥ 6; c ≥ 12. Find the greatest value of

3 42 6 12bc a ca b ab c

Sabc

− + ⋅ − + ⋅ −=

Solution

3 3

3 3 3

4 4

4 4

( 2) 22 ( 2).2

22 2 2 2

( 6) 3 36 ( 6).3.3

29 9 2. 9

( 12) 4 4 412 ( 12).4.4.4

464 64 8. 2

abc bc abcbc a a

bca ca abcca b b

cab ab abcab c c

� − +− = − ≤ ⋅ =

− + ++ ⋅ − = ⋅ − ≤ ⋅ =

− + + +⋅ − = ⋅ − ≤ ⋅ =

�

3 3

1 5 1

2 2 8 2 3. 9 8 2 2. 9

abc abc abcS

abc

� �� ≤ ⋅ + + = +� ��

� �

For

2 2

6 3

12 4

a

b

c

− =�

− =

− =�

⇔

4

9

16

a

b

c

=�

=

=�

, 3

5 1Max

8 2 3. 9S = +

Problem 5. Prove that: 3 3 1 12 11 ... 12 3

n nS nn

+ ++= + + + + < +

Solution

We have � 2

1factors

1 ( 1)1 11 1...1 1k

k

k

k kk k k

kk k k−

+ + −+ += ⋅ < = + . It follows

2 2 2

1 1 11 1 1 ... 1

2 3S

n

� � � � � �< + + + + + + +� � � � � �

� � � � � �2 2 2

1 1 1 1 1 1... ...

1 2 2 3 ( 1)2 3n n

n nn= + + + + < + + + +

× × − ×

( ) ( ) ( ) ( )1 1 1 1 1 1 1... 1 11 2 2 3 1

n n nn n n

= + − + − + + − = + − < +−

Problem 6. Prove that: 1 1 2n n

n nn n

n n+ + − <

Solution

�

�

2

2

1factors

1factors

1 ( 1)

1 1 . 1...1 1

1 ( 1)

1 1 . 1...1 1

n

nn nn

n

n

nn nn

n

n

n

nn

n nn n

n n n n

nn

n nn n

n n n n

−

−

� � �+ + − � �

� � � � + = + < = +� � � �+ � �

− + − � �� − = − < = −� � � ��

2 2

1 1 1 1 2n n n n

n nn n n n

n n n n

� � � �� + + − < + + − =� � � �

� � � �


20

3. POINT OF FREE INCIDENCE OR HOMOGENEOUS PRINCIPLE IN AM – GM INEQUALITY

Inequalities mentioned in this section are illustrated by polynomial functions for three

variables a, b, c > 0, which does not lose either the essence or the generality of the issue. We

will approach this technique from the simplest problems.

� Degree of the monomial: The monomial aαb

βc

γ has its degree as (α+β+γ)

For example: 3-degree monomials: 4 4 4 5 5 5

3 3 3 2 2 2

2 2 2, , , , , , , , , , ,

a b c a b ca b c a b b c c a

b c a b c a,

5 5 5 6 6 7

3 2 4, , , , ,

a b c a a a

bc ca ab b b c b,

7 7 8 8 8 9

3 2 2 5 3 2 4 4 2, , , , ,

a a a a a a

bc b c b b c bc b c⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

Thus there are countless 3-degree monomials written by variables a, b, c and generally we

have countless given k-degree monomials written by three variables a, b, c.

� Setting the matter: First of all, we will compare F(a, b, c) with G(a, b, c) as polynomials

of different degrees, in the following representative problem:

Problem: Prove that: ∀a, b, c > 0 we have 2000 2000 2000a b c a b c+ + ≥ + + (*)

Analysis: Suppose (*) is true, then let a = b = c > 0 we have

(*) ⇔ 3a2000

≥ 3a ⇔ a2000

≥ a

Since a2000

≥ a is true for a ≥ 1 and a2000

≥ a is false for a ∈ (0, 1), it follows

� If a, b, c > 0, 2000 2000 2000a b c a b c+ + ≥ + + is false

� If reducing the definite area a, b, c ≥ 1 then a2000

≥ a, b2000

≥ b, c2000

≥ c and we will have a

mediocre inequality 2000 2000 2000

a b c a b c+ + ≥ + + , ∀a, b, c ≥ 1

� If reducing the definite area a, b, c ∈(0,1) then a2000

< a, b2000

< b, c2000

< c and we will

also have a mediocre inequality 2000 2000 2000a b c a b c+ + < + + ∀a, b, c ∈ (0, 1)

� Conclusion: We should not compare polynomials of different degrees on the definite area

+� . Since there are infinite ways of writing given k-degree monomials, the following

problems set the matter that compares polynomial functions of same degree with degrees of

monomials written in different forms.

� General principle: When applying this technique we need to note that the monomials added

to use AM −−−− GM Inequality must have the same degrees with the monomials taken from the

polynomials in the inequality to be proved. This technique will be illustrated more clearly in

the following problems.

All inequalities following are generally in the form of fraction. Therefore, the most common

technique to solve them is eliminating denominators to turn them into the form of

polynomials. If we choose to eliminate denominators by equivalent transformation, i.e. equalizing

denominators, the solution will be very lengthy. Here, we will eliminate denominators by

adding reasonable expressions to use AM – GM inequality. All inequalities in the following

sections can be generated by the same ideas. However the generating idea is not so special so

it will be remained for the readers.


21

Problem 1. Prove that 2 2 2

, , 0a b c

a b c a b cb c a

+ + ≥ + + ∀ >

Proof

Remark: Both sides are polynomials containing 1-degree monomials

Using AM −−−− GM Inequality we have

2 22

2 22

2 22

2 2 2

2 2 2

2 2 2

a ab b a a

b b

b bc c b b

c c

c ca a c c

a a

� + ≥ ⋅ = =

+ + ≥ ⋅ = =

+ ≥ ⋅ = =�

2 2 2

( ) 2 ( )a b c

a b c a b cb c a

� �� + + + + + ≥ + +� �

� � � (q.e.d.)

Equality occurs ⇔ a = b = c > 0


2 2 2, , , 0

a b ca b c a b c

b c a+ + ≥ + + ∀ >

Proof



3 3

3

2 2

3 3

3

2 2

3 3

3

2 2

3. 3

3. 3

3. 3

a ab b b b a

b b

b bc c c c b

c c

c ca a a a c

a a

� + + ≥ ⋅ ⋅ =

+ + + ≥ ⋅ ⋅ =

+ + ≥ ⋅ ⋅ =�

3 3 3

2 2 22( ) 3( )

a b ca b c a b c

b c a

� �� + + + + + ≥ + +� �

� � � (q.e.d.)



22

Problem 3. Prove that 3 3 3 2 2 2

2 2 2, , 0

a b c a b ca b c

b c ab c a+ + ≥ + + ∀ >

Proof

� 1st solution: �Lemma:

2 2 2

, , 0a b c a b c a b cb c a

+ + ≥ + + ∀ >

��Application:

3 3 2

2 2

3 3 2

2 2

3 3 2

2 2

2 2

2 2

2 2

a a aa a

bb b

b b bb b

cc c

c c cc c

aa a

� + ≥ ⋅ =

+ + ≥ ⋅ = + ≥ ⋅ =�

3 3 3 2 2 2

2 2 2( ) 2a b c a b ca b c

b c ab c a

� � � �� + + + + + ≥ + +� ��

2 2 2

( )a b c

a b cb c a

� �≥ + + + + +� ��

3 3 3 2 2 2

2 2 2

a b c a b cb c ab c a

� + + ≥ + + . Equality occurs ⇔ a = b = c > 0

� 2nd

solution: � Lemma: 3 3 ( ) , 0x y xy x y x y+ ≥ + ∀ >

Proof:

33 3 3 3 3 3 2

33 3 3 3 3 3 2

3. 3

3. 3

x x y x x y x y

y y x y y x y x

� + + ≥ =+ + + ≥ =�

� 3 3 3 33( ) 3 ( ) ( )x y xy x y x y xy x y+ ≥ + � + ≥ +

Or proof : 3 3 2 2( )( ) ( )(2 ) ( )x y x y x y xy x y xy xy xy x y+ = + + − ≥ + − = +

Application: 3 3 3 2

2 2 2

3 3 3 2

2 2 2

3 3 3 2

2 2 2

( )

( )

( )

ab a ba a b ab a

bb b b

bc b cb b c bc b

cc c c

ca c ac c a ca c

aa a a

� +++ = ≥ = +

++

+ + = ≥ = +

++ + = ≥ = +�

3 3 3 2 2 2

2 2 2( ) ( )a b c a b ca b c a b c

b c ab c a

� � � �� + + + + + ≥ + + + + +� � � ��

3 3 3 2 2 2

2 2 2

a b c a b cb c ab c a

� + + ≥ + +



23


, , , 0a b c

a b c a b cbc ca ab

+ + ≥ + + ∀ > (Canada MO 2002)

Proof



3 33

3 33

3 33

3 3

3 3

3 3

a ab c b c a

bc bc

b bc a c a b

ca c

c ca b a b c

a a

� + + ≥ ⋅ ⋅ ⋅ =

+ + + ≥ ⋅ ⋅ ⋅ =

+ + ≥ ⋅ ⋅ ⋅ =�

� ( ) ( )3 3 3

2 3a b c

a b c a b cbc ca ab

+ + + + + ≥ + + � (q.e.d)



, , , 0a b c

ab bc ca a b cb c a

+ + ≥ + + ∀ >

Proof



3 3 3 33

3 3 3 33

3 3 3 33

3. 3

3. 3

3. 3

a b a bbc bc ab

b c b c

b c b cca ca bc

c a c a

c a c aab ab ca

a b a b

� + + ≥ ⋅ ⋅ =

+ + + ≥ ⋅ ⋅ =

+ + ≥ ⋅ ⋅ =�

3 3 3

2 ( ) 3( )a b c

ab bc ca ab bc cab c a

� �� + + + + + ≥ + +� �

� �

3 3 3

a b cab bc ca

b c a� + + ≥ + + .



24


2 2 2

3 3 3, , 0

a b ca b c a b c

b c a+ + ≥ + + ∀ >

Proof



5 5 5 52 2 2 2 2 2 25

3 3 3 3

5 5 5 52 2 2 2 2 2 25

3 3 3 3

5 5 5 52 2 2 2 2 2 25

3 3 3 3

5. 5

5. 5

5. 5

a a a ab b b b b b a

b b b b

b b b bc c c c c c b

c c c c

c c c ca a a a a a c

a a a a

� + + + + ≥ ⋅ ⋅ ⋅ ⋅ =

+ + + + + ≥ ⋅ ⋅ ⋅ ⋅ =

+ + + + ≥ ⋅ ⋅ ⋅ ⋅ =�

� ( ) ( )5 5 5

2 2 2 2 2 2

3 3 32 3 5

a b ca b c a b c

b c a

� �+ + + + + ≥ + +� �

� �

� 5 5 5

2 2 2

3 3 3

a b ca b c

b c a+ + ≥ + + . Equality occurs ⇔ 0a b c= = >


3 3 3 2 2 2, , , 0

a b c a b ca b c

b c a b c a+ + ≥ + + ∀ >

Proof



45 5 5 5 5 4

2 253 3 3 3 3 2

45 5 5 5 5 4

2 253 3 3 3 3 2

45 5 5 5 5 4

2 253 3 3 3 3 2

5. 5

5. 5

5. 5

a a a a a ab b

b b b b b b

b b b b b bc c

c c c c c c

c c c c c ca a

a a a a a a

�� + + + + ≥ ⋅ =� � � �

� �

+ + + + + ≥ ⋅ =� ��

� � + + + + ≥ ⋅ =� � � ��

5 5 5 4 4 4

2 2 2

3 3 3 2 2 24 ( ) 5 (1)

a b c a b ca b c

b c a b c a

� � � �� + + + + + ≥ + +� � � �

� � � �


25

On the other hand

4 42 2 4 2

2 2

4 42 2 4 2

2 2

4 42 2 4 2

2 2

2 2 2

2 2 2

2 2 2

a ab b a a

b b

b bc c b b

c c

c ca a c c

a a

� + ≥ ⋅ = =

+ + ≥ ⋅ = = + ≥ ⋅ = =�

4 4 42 2 2 2 2 2

2 2 2( ) 2( )

a b ca b c a b c

b c a

� �� + + + + + ≥ + +� �

� �

4 4 42 2 2

2 2 2

a b ca b c

b c a� + + ≥ + + (2). From (1) and (2) follows

5 5 5 4 4 4

2 2 2 2 2 2

3 3 3 2 2 24 ( ) 4 ( )

a b c a b ca b c a b c

b c a b c a

� � � �+ + + + + ≥ + + + + +� � � �

� � � �

⇔ 5 5 5 4 4 4

3 3 3 2 2 2

a b c a b c

b c a b c a+ + ≥ + + (q.e.d.).

Equality occurs ⇔ 0a b c= = >


3 3 3 2 2 2, , 0

a b c a b ca b c

b c a b c a+ + ≥ + + ∀ >

Proof



3 3 3 3 2

33 3 3 3 2

3 3 3 3 2

33 3 3 3 2

3 3 3 3 2

33 3 3 3 2

1 3. 1 3

1 3. 1 3

1 3. 1 3

a a a a a

b b b b b

b b b b b

c c c c c

c c c c c

a a a a a

�+ + ≥ ⋅ ⋅ = ⋅

+ + + ≥ ⋅ ⋅ = ⋅

+ + ≥ ⋅ ⋅ = ⋅�

3 3 3 2 2 2 2 2 2

3 3 3 2 2 2 2 2 22 3 2

a b c a b c a b c

b c a b c a b c a

� � � � � �� + + + ≥ + + + + +� � � � � �

� � � � � �

2 2 2 2 2 2

3

2 2 2 2 2 22 3.

a b c a b c

b c a b c a

� �≥ + + + ⋅ ⋅� �

� �

2 2 2

2 2 22 3

a b c

b c a

� �= + + +� �

� �

� 3 3 3 2 2 2

3 3 3 2 2 2

a b c a b c

b c a b c a+ + ≥ + + Equality occurs ⇔ 0a b c= = >


26


5 5 5 3 3 3

1 1 1, , 0

a b ca b c

b c a b c a+ + ≥ + + ∀ >

Proof

Remark: Both sides are polynomials containing (–3)-degree monomials

2 2 2 2 2 2

5

5 5 5 3 3 5 5 5 3 3 3

2 2 2 2 2 2

5

5 5 5 3 3 5 5 5 3 3 3

2 2 2 2 2 2

3

5 5 5 3 3 5 5 5 3 3 3

1 1 1 1 15. 5

1 1 1 1 15. 5

1 1 1 1 15. 5

a a a a a a

b b b a a b b b a a b

b b b b b b

c c c b b c c c b b c

c c c c c c

a a a c c a a a c c a

� + + + + ≥ ⋅ ⋅ ⋅ ⋅ = ⋅

+ + + + + ≥ ⋅ ⋅ ⋅ ⋅ = ⋅

+ + + + ≥ ⋅ ⋅ ⋅ ⋅ = ⋅�

2 2 2

5 5 5 3 3 3 3 3 3

1 1 1 1 1 13 2 5

a b c

b c a a b c a b c

� � � � � �� + + + + + ≥ + +� � � � � �

� � � ��

2 2 2

5 5 5 3 3 3

1 1 1a b c

b c a a b c� + + ≥ + + . Equality occurs ⇔ 0a b c= = >


1 1 1 1abca b abc b c abc c a abc

+ + ≤+ + + + + +

, ∀a, b, c > 0

(USAMO 1998)

Proof

Remark: ( ) ( ) ( ) ( ) ( )3 3 2 2 2 , , 0x y x y x y xy x y xy xy x y xy x y+ = + + − ≥ + − = + ∀ ≥

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

3 3

3 3

3 3

1 1 1

1 1 1

1 1 1

c

a b ab abc ab a b c abc a b ca b abc

a

b c bc abc bc a b c abc a b cb c abc

b

c a ca abc ca a b c abc a b cc a abc

� ≤ = = + + + + + ++ +

≤ = =+ + + + + + ++ +

≤ = =

+ + + + + ++ +�

� ( )3 3 3 3 3 3

1 1 1 1a b c

abcabc a b ca b abc b c abc c a abc

+ ++ + ≤ =+ ++ + + + + +

Problem 11. Prove that , , , 0a b c a b c a b cb c c a a b a b b c c a

+ + ≥ + + ∀ >+ + + + + +

Proof


( ) ( )a b c b c ab c c a a b a b b c c a

+ + + + ++ + + + + +

33 3b c c a a b b c c a a b

a b b c c a a b b c c a

+ + + + + += + + ≥ ⋅ ⋅ ⋅ =+ + + + + +

(1)


27

( ) ( ) 3a b c b c a a b b c c a

a b b c c a a b b c c a a b b c c a

+ + ++ + + + + = + + =+ + + + + + + + +

(2)

From (1) and (2) it follows: a b c a b cb c c a a b a b b c c a

+ + ≥ + ++ + + + + +

(q.e.d.)


Problem 12. Prove that 3 , , , 02

a b c a b cb c c a a b

+ + ≥ ∀ >+ + +

Proof

; ;a b c b c a c a bS A Bb c c a a b b c c a a b b c c a a b

= + + = + + = + ++ + + + + + + + +

3

3

3 3

3 3

a b b c c a a b b c c aA Sb c c a a b b c c a a b

c a a b b c c a a b b cB Sb c c a a b b c c a a b

+ + + + + ++ = + + ≥ ⋅ ⋅ ⋅ =+ + + + + +

+ + + + + ++ = + + ≥ ⋅ ⋅ ⋅ =+ + + + + +

3b c c a a bA Bb c c a a b

+ + ++ = + + =+ + +

� ( ) ( ) ( ) 36 2 3 2 3 22

A S B S A B S S S S≤ + + + = + + = + � ≤ � ≥

Problem 13. Prove that ( ) ( ) ( ) ( )3

21 1 1 2

a b ca b c

b c a abc

+ ++ + + ≥ + (1) (APMO 1998)

Proof

(1) ⇔ ( ) ( ) ( ) ( )

3 3

2 22 2

a b c a b ca b c b c a a b c b c a

b c a a b c b c a a b cabc abc

+ + + ++ + + + + + ≥ + ⇔ + + + + + ≥ (2)

Using AM −−−− GM Inequality we have:

( )

3

3

3

3

3

3

3

33

33

33

3

a a b a a b a

b b c b b c abc

b b c b b c b

c c a c c a abc

c c a c c a c

a a b a a b abc

a b c a b c

b c a abc

�+ + ≥ ⋅ ⋅ ⋅ =

+ + + ≥ ⋅ ⋅ ⋅ = + + ≥ ⋅ ⋅ ⋅ =�

+ +� + + ≥

( )

3

3

3

3

3

3

3

33

33

33

4

b b c b b c b

a a b a a b abc

c c a c c a c

b b c b b c abc

a a b a a b a

c c a c c a abc

b c a a b c

a b c abc

�+ + ≥ ⋅ ⋅ ⋅ =

+ + + ≥ ⋅ ⋅ ⋅ = + + ≥ ⋅ ⋅ ⋅ =�

+ +� + + ≥

From (3) and (4) it follows q.e.d.


28

Problem 14. Prove that 7 7 7 5 2 5 2 5 2 , , , 0a b c a b b c c a a b c+ + ≥ + + ∀ >

Proof


77 7 7 7 7 7 7 7 7 7 7 7 7 7 5 2

77 7 7 7 7 7 7 7 7 7 7 7 7 7 5 2

77 7 7 7 7 7 7 7 7 7 7 7 7 7 5 2

7 7

7 7

7 7

a a a a a b b a a a a a b b a b

b b b b b c c b b b b b c c b c

c c c c c a a c c c c c a a c a

� + + + + + + ≥ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =

+ + + + + + + ≥ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =

+ + + + + + ≥ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =�

� ( ) ( )7 7 7 5 2 5 2 5 27 7a b c a b b c c a+ + ≥ + + � (q.e.d)

Problem 15. Prove that ( ) , , , 0;3 3

nn n na b c a b c a b c n+ + + +≥ ∀ > ∈�

Proof


( ) ( ) ( )( )

( )

( ) ( ) ( )( )

( )

( ) ( ) ( )( )

( )

1 1

1 1

1 1

13 3 3

13 3 3

13 3 3

n n n nn nn

n n n nn nn

n n n nn nn

a b c a b c a b ca n n a n a

a b c a b c a b cb n n b n b

a b c a b c a b cc n n c n c

− −

− −

− −

�+ + + + + ++ − ≥ ⋅ =

+ + + + + ++ + − ≥ ⋅ = + + + + + ++ − ≥ ⋅ =�

( ) ( ) ( ) ( ) ( ) ( )1

3 1 33 3 3

n n nn n n a b c a b c a b ca b c n a b c n

−+ + + + + +� + + ≥ + + − − = � (q.e.d)

Problem 16. Prove that ( ) ( ) ( )2 2 2 , , , 0;3 3 3

n n nn n n a b b c c aa b c a b c n+ + ++ + ≥ + + ∀ > ∈�

Proof

� Lemma: , , , 0;3 3

nn n nx y z x y z

x y z n+ + + +� �

≥ ∀ > ∈� ��

�

� Application: Using lemma we have:

( ) ( )

( ) ( )

( ) ( )

23 3 3

23 3 3

23 3 3

n nn n n

n nn n n

n nn n n

a b b a b b a b

b c c b c c b c

c a a c a a c a

� + + + + +≥ = + + + + ++ ≥ =

+ + + + + ≥ =�

� ( ) ( ) ( )2 2 23 3 3

n n nn n n a b b c c aa b c + + ++ + ≥ + + (q.e.d)


29

Problem 17. Let be given a, b, c, d > 0. Find the minimum value of the expression:

S = a b c d

b c d c d a a b b a b c+ + + +

+ + + + + + + +

+ b c d c a d d a b a b c

a b c d

+ + + + + + + ++ + +

Solution

• Common mistake:

Using AM −−−− GM Inequality directly for eight factors

88 . . . . . . . 8a b c d b c d c d a d a b a b c

Sb c d c d a d a b a b c a b c d

+ + + + + + + +≥ ⋅ =

+ + + + + + + + � Min S = 8

• Causes of the mistakes:

Min S = 8 ⇔

a b c d

b c d a

c d a b

d a b c

= + +� = + +

= + +

= + +�

� ( )3a b c d a b c d+ + + = + + + � 1 = 3 : illogical

• Analysis and solutions:

To find Min S we need to note that S is a symmetric expression for a, b, c, d; thus, Min S

(or Max S) will occur at the "free point of incidence": 0a b c d= = = > .

Thus let be given a = b = c = d > 0 and estimate Min S = 4 1

12 133 3

+ =

It follows that the estimated condition for equality to occur in all the component inequalities

is the sub-set of the estimated condition 0a b c d= = = >

• Point of incidence Pattern: Let be given a = b = c = d > 0 we have

1

3

3

a b c d

b c d c d a d a b a b c


a b c d

�= = = = + + + + + + + +

+ + + + + + + +

= = = = α α α α α�

� 1 3

3=

α � α = 9

�� Right solution: Transforming and using AM −−−− GM Inequality we have

S = 8

.9 9 9cyc cyc

a b c d b c d

b c d a a

+ + + +� �+ + ≥� �

+ +� ��

≥ 88 . . . . . . .9 9 9 9

a b c d b c d c d a d a b a b c

b c d c d a d a b a b c a b c d

+ + + + + + + +⋅

+ + + + + + + +

+ 8

9


a a a b b b c c c d d d

� �+ + + + + + + + + + +� �

� �


30

≥ 128 8

123 9


a a a b b b c c c d d d+ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ =

8 8 8 32 40 1.12 13

3 9 3 3 3 3+ = + = =

For a = b = c = d > 0, Min S = 1

133

Problem 18. Let be given a, b, c, d > 0. Find the minimum value of the expression:

S = 2 2 2 2

1 1 1 13 3 3 3

a b c d

b c d a

� � � � � � � �+ + + +� � � � � � � �

� � � � � � � �

Solution

• Common mistake:

S = 2 2 2 2

1 1 1 13 3 3 3

a b c d

b c d a

� �� + + + +� ��

� �� ≥

2 2 2 2 642 2 2 2

3 3 3 3 9

a b c d

b c d a⋅ ⋅ ⋅ = �

64Min

9S =

• Causes of the mistake:

64Min

9S = ⇔

2 2 2 21

3 3 3 3

a b c d

b c d a= = = = =

2( ) 2

3( ) 3

a b c d

a b c d

+ + +=

+ + + � illogical

• Analysis and solutions:

Since S is a symmetric expression for a, b, c, d we estimate Min S occurs at Free point of

incidence: a = b = c = d > 0, then S =

42 625

13 81

� �+ =� �

� �

� Right solution: Using AM −−−− GM Inequality we have

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

23 2

55

23 2

55

23 2

55

23 2

55

2 1 1 1 511 53 33 3 3 3 3 3 3

2 1 1 1 511 53 33 3 3 3 3 3 3

2 1 1 1 511 53 33 3 3 3 3 3 3

2 1 1 1 511 53 33 3 3 3 3 3 3

a a a a a

b bb b b

b b b b b

c cc c c

c c c c c

d dd d d

d d d d d

a aa a a

� + = + + + + ≥ ⋅ ⋅ =

+ = + + + + ≥ ⋅ ⋅ =+

+ = + + + + ≥ ⋅ ⋅ = + = + + + + ≥ ⋅ ⋅ =�

� S = 2 2 2 2

1 1 1 13 3 3 3

a b c d

b c d a

� � � � � � � �+ + + +� � � � � � � �

� � � � � � � � ≥

2

5625 625

81 81

a b c d

b c d a

� �⋅ ⋅ ⋅ ⋅ =� ��

For a = b = c = d > 0, Min S = 625

81


31

4. METHOD OF SPECIALIZING INEQUALITY OF THE SAME DEGREE:

Problem 1. Given , , 0

3

a b c

a b c

>�

+ + =�. Prove that

( ) ( ) ( ) ( ) ( ) ( )

3 3 3 34

a b c

a b a c b c b a c a c b+ + ≥

+ + + + + +

Solution

We can transform the above inequality to the 1-degree homogeneous inequality as follows:

( ) ( ) ( ) ( ) ( ) ( )

3 3 3

4a b c a b c

a b a c b c b a c a c b

+ ++ + ≥+ + + + + +

Using AM – GM inequality, we have:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

3 33

3 33

3 33

338 8 8 8 4

338 8 8 8 4

338 8 8 8 4

a a b a c a a b a c a

a b a c a b a c

b b c b a b b c b a b

b c b a b c b a

c c a c b c c a c b c

c a c b c a c b

� + + + ++ + ≥ ⋅ ⋅ ⋅ =+ + + +

+ + + ++ + + ≥ ⋅ ⋅ ⋅ =

+ + + + + + + ++ + ≥ ⋅ ⋅ ⋅ = + + + +�

� ( ) ( ) ( ) ( ) ( ) ( )

3 3 3 34 4

a b c a b c


+ ++ + ≥ =+ + + + + +

(q.e.d)


3

a b c

a b c

>�


( ) ( ) ( )

3 3 3

12 2 2

a b c

b c a c a b a b c+ + ≥

+ + +

Solution


( ) ( ) ( )

3 3 3

32 2 2

a b c a b c

b c a c a b a b c

+ ++ + ≥+ + +

. Using AM – GM Inequality we have:

( )( )

( )( )

( )( )

( )( )

( )( )

( )( )

3 33

3 33

3 33

9 93 2 3 3 2 92 2

9 3 2 3 3 2 92 2

9 3 2 3 3 2 92 2

a ab c a b c a ab c a b c a

b bc a b c a b bc a b c a b

c ca b c a b c ca b c a b c

�+ + + ≥ ⋅ ⋅ + =

+ +

+ + + + ≥ ⋅ ⋅ + =+ +

+ + + ≥ ⋅ ⋅ + = + +�

� ( ) ( ) ( )

( ) ( )3 3 3

9 6 92 2 2

a b c a b c a b cb c a c a b a b c

� + + + + + ≥ + +� �+ + +� �

� ( ) ( ) ( )

3 3 3

132 2 2

a b c a b c

b c a c a b a b c

+ ++ + ≥ =+ + +

(q.e.d)


32


, , 0

1

a b c

a b c

>�

+ + =�

. Prove that 3 3 3

12 2 2 3

a b c

b c c a a b+ + ≥

+ + +

Solution


3 3 3 2 2 2

2 2 2 3a b c a b c

b c c a a b

+ ++ + ≥+ + +

Using AM – GM Inequality we have:

( ) ( )

( ) ( )

( ) ( )

3 32

3 32

3 32

9 92 2 2 62 2

9 92 2 2 62 2

9 92 2 2 62 2

a aa b c a b c ab c b c

b bb c a b c a bc a c a

c cc a b c a b ca b a b

�+ + ≥ ⋅ + =

+ +

+ + + ≥ ⋅ + =+ +

+ + ≥ ⋅ + = + +�

� ( ) ( )3 3 3

2 2 29 3 62 2 2

a b c ab bc ca a b cb c c a a b

� �+ + + + + ≥ + +� �

+ + +� �

( )3 3 3

2 2 29 32 2 2


� �� + + ≥ + +� �

+ + +� �

� 3 3 3 2 2 2

12 2 2 3 3

a b c a b c

b c c a a b

+ ++ + ≥ =+ + +


1

a b c

ab bc ca

>�


( ) ( ) ( )91 1 12a a b b b c c c a

+ + ≥+ + +

Solution


( )( )

( )( )

( )( )

92

c a b ab a b c bc b c a ca

a a b b b c c c a

+ + + + + ++ + ≥

+ + +

⇔ ( ) ( ) 92

a b c b c a

b c a a b b c c a+ + + + + ≥

+ + +

⇔ ( ) ( ) 152

a b b c c a b c a

b c a a b b c c a

+ + ++ + + + + ≥+ + +

(1)

We have: LHS (1) = ( )34 4 4 4

a b b c c a b c a a b b c c a

b c a a b b c c a b c a

+ + + + + ++ + + + + + + ++ + +

6 33 156 3 34 4 4 4 2

a b b c c a b c a a b c

b c a a b b c c a b c a

� �+ + +≥ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ + =� �+ + + � �


33


1

a b c

a b c

>�


( ) ( ) ( )2 2 2

94

a b c

b c c a a b+ + ≥

+ + +

Solution

� Lemma: 3 , , , 02


+ + ≥ ∀ >+ + +

( )22 2 2 1 , , ,

3x y z x y z x y z+ + ≥ + + ∀ ∈�

� Application:


( )

( )

( )

( )

( )

( )2 2 2

94

a a b c b a b c c a b c

b c c a a b

+ + + + + ++ + ≥

+ + +

( ) ( ) ( )2 2 2

94

a b c a b c

b c c a a b b c c a a b⇔ + + + + + ≥

+ + + + + + (1)

We have: LHS (1) ≥ ( ) ( )2 2

3 3 91 13 3 2 2 4

a b c a b c

b c c a a b b c c a a b+ + + + + ≥ + =

+ + + + + + � (q.e.d)


, , 0

3

a b c

a b c

>�

+ + =�

. Prove that 3ab bc ca

c a b+ + ≥ (1)

(France Pre-MO 2005)

Solution

(1) ⇔ ( ) ( ) ( )2 2 2 2 2 2 2

2 2 2 2 2 2

2 2 29 2 3ab bc ca a b b c c a a b c a b c

c a b c a b+ + ≥ ⇔ + + + + + ≥ + +

⇔ 2 2 2 2 2 2

2 2 2

2 2 2

a b b c c a a b cc a b

+ + ≥ + + . Using AM – GM Inequality we have:

2 2 2 2 2 2 2 22

2 2 2 2

2 2 2 2 2 2 2 22

2 2 2 2

2 2 2 2 2 2 2 22

2 2 2 2

2 2

2 2

2 2

a b b c a b b c bc a c a

b c c a b c c a ca b a b

c a a b c a a b ab c b c

�+ ≥ ⋅ =

+ + ≥ ⋅ =

+ ≥ ⋅ =�

� 2 2 2 2 2 2

2 2 2

2 2 2

a b b c c a a b cc a b

+ + ≥ + + � (q.e.d)


34


1

a b c

abc

>�

=�. Prove that

( ) ( ) ( ) ( ) ( ) ( )

3 3 3 341 1 1 1 1 1

a b c

b c c a a b+ + ≥

+ + + + + +

(IMO Shortlist 1998)

Solution


( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

3 33

3 33

3 33

1 1 1 1 338 8 8 8 41 1 1 1

1 1 1 1 338 8 8 8 41 1 1 1

1 1 1 1 338 8 8 8 41 1 1 1

a b c a b c a

b c b c

b c a b c a b

c a c a

c a b c a b c

a b a b

� + + + ++ + ≥ ⋅ ⋅ ⋅ =+ + + +

+ + + ++ + + ≥ ⋅ ⋅ ⋅ =

+ + + + + + + ++ + ≥ ⋅ ⋅ ⋅ = + + + +�

� ( ) ( ) ( ) ( ) ( ) ( )

33 3 3 33 3 32 4 2 4 4

abca b c a b c


⋅+ ++ + ≥ − ≥ − =+ + + + + +


1

a b c

abc

>�

=�. Prove that

( ) ( ) ( )3 3 3

31 1 12a b c b c a c a b

+ + ≥+ + +

(1)

(IMO 1995)

Solution

We can transform the above inequality to a stronger homogeneous one as follows:

( ) ( ) ( ) ( ) ( )2 2 2

3 3 3

1 1 1

1 1 1 1 1 1 1 12 1 1 1 1 1 1 2

abc abc abc a b c

a b c a b ca b c b c a c a bb c c a a b

+ + ≥ + + ⇔ + + ≥ + ++ + + + + +

(2)

Letting 1 1 1, ,x y za b c

= = = for , , 0x y z > , 1xyz = .

(2) ⇔ ( )22 2 1

2

yx z x y zy z z x x y

+ + ≥ + ++ + +

. Using AM – GM Inequality we have:

2 2

2 2

2 2

24 4

24 4

24 4

y z y zx x xy z y z

y yz x z x yz x z x

x y x yz z zx y x y

� + ++ ≥ ⋅ ⋅ =

+ + + ++ + ≥ ⋅ ⋅ =

+ +

+ ++ ≥ ⋅ ⋅ = + +�

� ( )22 2

3 31 1 32 2 2

yx z x y z xyzy z z x x y

+ + ≥ + + ≥ ⋅ ⋅ =+ + +

(q.e.d)


35

Problem 9. Let , , 0

1

a b c

abc

>�

=�. Prove that

5 5 5 5 5 51ab bc ca

a b ab b c bc c a ca+ + ≤

+ + + + + +


Solution

� Lemma: ( )5 5 3 2 2 3 2 2 , , 0x y x y x y x y x y x y+ ≥ + = + ∀ >

� Application: Using Lemma we have

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

5 5 2 2

5 5 2 2

5 5 2 2

1

1

1

1

1

1

ab ab abc c

a b cab a b ab a b abca b ab a b a b ab

bc bc abc a

a b cbc b c bc b c abcb c bc b c b c bc

ca ca abc b

a b cca c a ca c a abcc a ca c a c a ca

� ≤ = = = + ++ + + ++ + + ++ ≤ = = = + ++ + + ++ + + + ≤ = = =

+ ++ + + + + + + +�

� 5 5 5 5 5 5

1ab bc ca a b c

a b ca b ab b c bc c a ca

+ ++ + ≤ =+ ++ + + + + +


1

a b c

abc

>�

=�. Prove that

( ) ( ) ( )2 2 2

22 2

a b c b c a c a b

b b c c c c a a a a b b

+ + ++ + ≥

+ + + (1)

Solution

( )

( )

( )

22

22

22

.2 2 222 2 2

2.2 2

22 2 2

.2 2 22

a bc a aa b c x

y zb b c c b b c c b b c c

yb ca b bb c a

z xc c a a c c a a c c a a

c ab c cc a b zx ya a b b a a b b a a b b

� +≥ = = ++ + +

+

+ ≥ = =++ + +

+

≥ = = ++ + +�

in which

0

0

0

1

x a a

y b b

z c c

xyz

� = > = > = > =�

� S = ( ) ( ) ( )2 2 2 22 2

2 2 22 2

ya b c b c a c a b x zy z z x x yb b c c c c a a a a b b

+ + ++ + ≥ + +

+ + ++ + + = T

Letting

4 22 9

4 22

94 22

9

n p mxy z m

p m nz x n y

m n px y p z

+ −�=+ =�

+ −+ = � =

+ −+ =� =

�

� 4 2 4 2 4 22

9

n p m p m n m n pT

m n p

+ − + − + −� �= + +� �

� �

( )3 32 2 2

4 6 4 3 3 6 12 3 6 29 9 9

p p p pn m m n n m m n

m p n m n p m p n m n p

� �� = + + + + + − ≥ ⋅ ⋅ ⋅ ⋅ + ⋅ ⋅ ⋅ − = + − =� ��

� � � ��

Equality occurs ⇔ m n p= = ⇔ 1x y z= = = ⇔ 1a b c= = =


36

5. NON – SYMMETRIC POINT OF INCIDENCE IN AM – GM INEQUALITY

The following problems express essence of the term “point of incidence” correctly; means

that we can find any “point of incidence” and build up an inequality that reaches it extremum

at this point. Since these point are not symmetric so the readers must speculate the point of

incidence base on the condition and expression structure given.


2 3 20

a b c

a b c

>�

+ + ≥�. Prove that S =

3 9 413

2a b c

a b c+ + + + + ≥

Proof

Predict S = 1 at the point: a = 2, b = 3, c = 4

Applying AM – GM inequality, we have:

4 42 4

9 92 6

16 162 8

a aa a

b bb b

c cc c

�+ ≥ ⋅ =

+ ≥ ⋅ = + ≥ ⋅ =�

�

( )

( )

( )

3 34 4 34 4

1 19 6 32 2

1 116 8 24 4

aa

bb

cc

�+ ≥ ⋅ =

+ ≥ ⋅ = + ≥ ⋅ =�

3 1 1 3 9 4

84 2 4 2

a b ca b c

� + + + + + ≥ (1)

We have a + 2b + 3c ≥ 20 1 3

54 2 4

ba c� + + ≥ (2)

Add (1) and (2) � S = 3 9 4

132

a b ca b c

+ + + + + ≥

Problem 2. Given a, b, c > 0. Prove that

S = ( )3

2 2 1 1 130 3 36 849

ca b

ab bc ca+ + + + + ≥

Proof

Predict S = 84 at the point: a = 1, b = 2, c = 3


( )

( )

( )

2 2225

3 22 3 62 311

3 3337

2 22. 1 2 5 544

6 63 2 6 11 114 274 27

3 33. 3 7 72727

b ba aabab

b c b c

bcbc

c ca acaca

�+ ⋅ + ⋅ ≥ ⋅ ⋅ =

� � � � ⋅ + ⋅ + ⋅ ≥ ⋅ =� � � �

� � � �

+ + ⋅ ≥ ⋅ ⋅ =�

2

2 3

3

49 2 45

4

3 2 3611

4 27

94 3 28

27

ba

ab

b c

bc

ca

ca

� � �+ + ≥ � �

� �

� + + ≥� �

+ + ≥� � � ��

� 3

2 2 1 1 130 3 36 45 11 28 84

9

ca b

ab bc ca

� �+ + + + + ≥ + + ≥� �

� �


37


12 ; 8

a b c

ac bc

>�

≥ ≥�. Prove that S = ( )

1 1 1 8 1212

2a b c

ab bc ca abc

� �+ + + + + + ≥� �

� �

Proof

Predict S = 121

2 at the point: a = 3, b = 2, c = 4


3

3

3

4

6 63 3

3 2 3 2

8 83 3

2 4 2 4

12 123 3

4 3 4 3

24 244 4

3 2 4 3 2 4

a b a b

ab ab

b c b c

bc bc

c a c a

ca ca

a b c a b c

abc abc

�+ + ≥ ⋅ ⋅ ⋅ =

+ + ≥ ⋅ ⋅ ⋅ =

+ + ≥ ⋅ ⋅ ⋅ = + + + ≥ ⋅ ⋅ ⋅ ⋅ =�

�

61 3

3 2

84 12

2 4

127 21

4 3

241 4

3 2 4

a b

ab

b c

bc

c a

ca

a b c

abc

� � �⋅ + + ≥� ��

� � ⋅ + + ≥� � � �

� �

⋅ + + ≥� ��

� � ⋅ + + + ≥� � � ��

( )6 32 84 24

3 40a b cab bc ca abc

+ + + + + + ≥ . Also 12

8

ac

bc

≥�

≥� therefore

1 112; 8

ac bc≤ ≤

Hence 1 140 3 26 78Sbc ca

≤ + ⋅ + ⋅1 1

3 26 788 12

S≤ + ⋅ + ⋅

363

32

S� ≤ 121

2S⇔ ≤ (q.e.d)

Problem 4. Let be given , , 0 satisfying 1a b c a b c> + + = . Prove that

S = 4 4 4 256

3125a b b c c a+ + ≤

Proof

WLOG, supposing { } 4 3Max , ,a a b c b c a bc= � ≤ and 4 2 3 4c a c a ca≤ ≤ .

Since 3

4 2

c c≥ then transform the expression S we have

4 4 4 2 34 4 4 3

2 2 2 2

c a c a ca c aS a b b c a b a bc= + + + ≤ + + + ( ) ( ) ( ) ( )

33 3

22

a c ca b a c a c a a c b= + + + = + +

� ( ) ( ) ( ) ( )4 4 4 3 3 32 2c cS a b b c c a a a c b a a c b= + + ≤ + + ≤ + +


38

S = ( )( )

5

4 4

323 4 4 4 44 4

24 4 4 4 5

a a a a c cba a a a c cb

+� + + + + +� �+

⋅ ⋅ ⋅ + ≤ � ��

=

5 44

5

4 2564 .

5 31255

a b c+ +� �= =� �

� � �

4 4 4 256

3125S a b b c c a= + + ≤

Equality occurs (corresponding to a = Max (a, b, c)) ⇔

1; 04 1

; ; 03 5 54 4 4

a b c c

a b ca a c cb

+ + = =�

⇔ = = = += = +

�

In the general case the inequality becomes equality ⇔ (a, b, c) is the permutation ( )4 1, , 05 5

Problem 5. Given 0 1a b c≤ ≤ ≤ ≤ . Prove that: 2 2 2 108( ) ( ) (1 )

529a b c b c b c c− + − + − ≤

Proof

Transforming and using AM – GM Inequality we have

( )2 2 2 21( ) ( ) (1 ) 0 . 2 2 (1 )

2a b c b c b c c b b c b c c� − + − + − ≤ + − + −� �

≤ ( )3

2 2 21 2 2 4 231 1 1

2 3 27 27

b b c bc c c c c c c

+ + −� � � � � �⋅ + − = ⋅ + − = − ⋅� � � � � ��

=

2 2 354 23 23 23 54 1 108

123 54 54 27 23 3 529

c c c� � � � � � � � � � � �

⋅ ⋅ − ⋅ ≤ ⋅ =� � � � � � � � � � � ��

Equality occurs ⇔ a = 0, b = 12

23, c =

18

23

Problem 6. Let x, y, z, t ∈[0, 1]. Find the minimum value of

S = 2 2 2 2 2 2 2 2

x y y z z t t x xy yz zt tx+ + + − − − −

Solution

WLOG suppose { }Max , , ,x x y z t= . Then we have

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2S y x z yz xy t z x xt zt y x z x z y t z x x z t= − + − + − + − = − + − + − + −

( ) ( )( ) ( )

3

3 38 8 80 1

3 27 27 27

y x z x z yy x z x z y x

� + − + + −≤ − + − + ≤ = ⋅ ≤ ⋅ =� �

� �

For 2 11; ; ; 03 3

x y z t= = = = , 8

Max27

S =


39

Problem 7. Let be given , , 0a b c ≥ satisfying 3a b c+ + = . Prove that

( ) ( ) ( )2 2 2 2 2 2 12a ab b b bc c c ca a− + − + − + ≤ (1)

Proof

WLOG supposing 0a b c≥ ≥ ≥ �

2 2 2

2 2 2

0

0

b bc c b

c ca a a

� ≤ − + ≤ ≤ − + ≤�

� ( ) ( )2 2 2 2 2 2b bc c c ca a a b− + − + ≤

LHS (1) ( ) ( ) ( )( )3

2 2 2 2 2 2 2 23 3 3 34 4 19 2 2 9 3 2 2

ab ab ab aba b a ab b a ab b a ab b� ≤ − + = ⋅ ⋅ − + ≤ + + − +� ��

( ) ( ) ( )

3 32 2 6 6

3

3 3

34 4 4 4 4 3 129 3 9 3 9 9 93 3

a b a b c a b c� � � �+ + + + += ≤ = ⋅ = ⋅ = ⋅ =� � � �

� � � �

Equality occurs ⇔ 2 23 , 0 2, 1, 02ab a ab b c a b c= − + = ⇔ = = =

Problem 8. Given 7; 5 7 70

10 14 35 210

a a b

a b c

≥ + ≥�

+ + ≥�. Find the minimum value of S =

3 3 3a b c+ +

Solution

� Lemma:

33 3 3

3 3

x y z x y z+ + + +� �≥ � ��

∀x, y , z > 0

� Proof:

3 3 2

3

3 3 2

3

3 3 2

3

33 3 3

33 3 3

33 3 3

x y z x y z x y zx x

x y z x y z x y zy y

x y z x y z x y zz z

� + + + + + +� � � � � �+ + ≥ � � � � � ��

+ + + + + +� � � � � �+ + + ≥� � � � � �

� � � � � � + + + + + +� � � � � �

+ + ≥� � � � � � � � � � � ��

� ( )3 2 3

3 3 3 3 3 36 3 33 3 3

x y z x y z x y zx y z x y z x y z

+ + + + + +� � � � � �+ + + ≥ + + ⇔ + + ≥� � � � � �

� � � � � �

� Application:

3 3 3 3 3 3

3 3 3 8 117 2187 5 2 7 5 7

a b c a b aS a b c

� � � � � � � � � � � � � �= + + = + + + + +� � � ��

� ��

3 3

3

7 5 2 7 58 3 117 2 218 8 3 117 2 218 4763 2 7

a b c a b

a

� � � � � �+ + +� � � �� ≥ + + ≥ × + × + =� � � � � �

� ��

With a = 7, b = 5, c = 2 we have Min S = 476


40

6. METHOD OF EQUALIZING COEFFICIENTS:

In five sections above, we can predict the point of incidence of inequalities based on the

conditions given in the assumption and mathematical features in the expressions. However, if

the point of incidence is not that easy to be predicted, it is necessary to find out a method that

can help define the exact point. This is also the content of balancing coefficient technique.

Problem 1. Let x ∈ [0, 1]. Find the maximum value of 2 4 2 413 9x x x x= − + +�

Solution

For , 0α β > , using AM – GM inequality we have:

( )( ) ( )

( )( ) ( )

2 2 2 2 22 4 2 2 2

2 22 2 22 4 2 2 2

1 13 1 1313 1313 122

9 1 919 99 122

x x xx x x x

xx xx x x x

� α + − α − +− = ⋅ α − ≤ ⋅ =

α α α+ β + +β + +

+ = ⋅ β + ≤ ⋅ =β β β�

� ( ) ( )22

2 4 2 4 29 113 1 13 913 92 2 2 2

S x x x x x� β +α −

= − + + ≤ + + +� �α β α β� �

Equality occurs ⇔ ( ) ( )2 2 2

2 2 2 2

2 2 2

11 1 1

1

x xx x

x x

�α = −⇔ α + = β − =

β = +�

Choose , 0α β > such that ( ) ( )

2 2

22

1 1

31 ;9 1 2 213 1

02 2

�α + = β −

⇔ α = β = β +α − + =α β�

. Then 13 9 162 2

S ≤ + =α β

Equality occurs ⇔ ( )2 21 1xα + = ⇔ 2 2 55 14 5

x x= ⇔ = . Thus Max S = 16

Problem 2. Let be given a > 0 and 2 2 2 2916

x y z xy a+ + + = .

Find the maximum value of S xy yz zx= + +

Solution

For ( )0;1α∈ , using AM – GM inequality we have:

( )

( ) ( ) ( )

( ) ( ) ( )

2 2

2 22 2

2 22 2

2 .

1 2 1 2 1 .2 2

1 2 1 2 1 .2 2

x y xy

z zx x xz

z zy y yz

�α + ≥ α − α + ≥ − α ⋅ ≥ − α+

− α + ≥ − α ⋅ ≥ − α�

� ( ) ( )2 2 2 2 . 2 1x y z xy xz yz+ + ≥ α + − α + � ( ) ( ) ( )2 92 2 116

a xy xz yz≥ α + + − α +


41

Choose ( )0;1α∈ such that ( ) 12 5 1792 2 116 32

−α + = − α ⇔ α = .

Then we have ( )2 3 5 2

4a xy yz zx

−≥ + + �

( )224 3 5 24

413 5 2

aS xy yz zx a+

= + + ≤ =−

� ( ) 24 3 5 2

Max41

S a+

=

Equality occurs ⇔

2 2 2 2

49 12 54 ;49 12 5 49 12 5 290 12 590 12 532 32 2

9 49 12 54 ;16 90 12 590 12 5

ax y z ax y z

ax y z xy a x y z a

� −� = = =�− −= = −� − �⇔

� −−+ + + = � = = = −� � −−�


0

2 3 1

a,b,c

a b c

≥�

+ + =�

Find the minimum value of S = 3 3 32 3 4a b c+ +

Solution

Supposing , , 0α β γ ≥ and using AM – GM inequality, we have:

( )

( )

3 3 3 2

3 3 3 2

3 3 3 2

3

3 92 2

2 6

a a a

b b b

c c c

� + + α ≥ α

+ + + β ≥ β

+ + γ ≥ γ�

� 3 3 3 3 3 3 2 2 23 92 3 4 2 3 . 2 2 32 4

a b c a b c+ + + α + β + γ ≥ α + β ⋅ + γ ⋅

Equality occurs ⇔ 2 2 2

, ,

2 3 1

a b c

a b c

= α = β = γ�

+ + =�

Choose , ,α β γ such that

( ) 22 2 2

93 2 093 2 0 4 9 184 3 24 40732 31 12 3 1

9 81 4

kkk

k

� α = β = γ = >� α = β = γ = ≥ ⇔ ⇔ α = β = γ = =

+ + =α + β + γ =� �

Then we have: ( )3 3 3

3 3 3 2 2 26 3 8 9 182 3 4 2 2 32407 407 407 407

a b c a b c� � � � � �+ + + + + ≥ + +� � � � � ��

⇔ 3 3 3 3 3 36 18 122 3 4 2 3 4

407 407 407a b c a b c+ + + ≥ ⇔ + + ≥

For 6 8 9, ,407 407 407

a b c= = = , we have 12Min407

S =


42

Problem 4. Given 0

3

a,b,c

a b c

≥�

+ + =� . Find the maximum value of S = 4 8 6ab bc ca+ +

Solution

Assume that ( ) ( ) ( ) ( ) ( ) ( )S ma b c nb c a pc a b m n ab n p bc p m ca= + + + + + = + + + + +

�

4 1

8 3

6 5

m n m

n p n

p m p

+ = =� �

+ = ⇔ = + = =� �

� ( ) ( ) ( )

( ) ( ) ( )

4 8 6 3 5

3 3 3 5 3

S ab bc ca a b c b c a c a b

a a b b c c

= + + = + + + + +

= − + − + −

� ( ) ( ) ( )2 2 2

81 3 3 33 54 2 2 2

S a b c�

= − − + − + −� ��

Take 3 3 3; ;2 2 2

x a y b z c= − = − = − � 9 32 2

x y z a b c+ + ≥ + + − =

Then: ( )2 2 2 2 2 281 813 5 3 54 4

S x y x x y x S= − + + ⇔ + + = − . With α, β, γ > 0 we have

2 2 2 2

2 2 2 2

2 2 2 2

2 2

3 3 6 6

5 5 10 10

x x x

y y y

z z y

� + α ≥ α = α

+ + β ≥ β = β

+ γ ≥ γ = γ�

� ( ) ( ) ( )2 2 281 3 5 2 3 54

S x y z− + α + β + γ ≥ α + β + γ

Choose 3 5α = β = γ

� ( ) ( ) ( )2 2 2 2 2 281 813 5 2 3 5 34 4

S x y z≤ + α + β + γ − α + + ≤ + α + β + γ − α

( )2 2 281Max 3 5 34

S = + α + β + γ − α ⇔ 3; ; và2

x y z x y z= α = β = γ + + =

3 45 15 9; ;3 5 2 46 46 46

x y z α α� + + = α + β + γ = α + + = � α = β = γ = . Hence:

( )2

2 2 281 81 23 432Max 3 5 3 34 4 15 23

S α= + α + β + γ − α = + − α =

The equality occurs ⇔ 3 45 3 15 3 9; ;2 46 2 46 2 46

a b c− = α = − = β = − = γ =

⇔ 3 45 3 15 3 9 27 3012; ; ; ;2 46 2 46 2 46 23 23 23

a b c a b c− = − − = − − = − ⇔ = = =

Problem 5. Let , , , 0a b c m > be satisfying 1ab bc ca+ + = .

Find the minimum value of ( )2 2 2S m a b c= + + according to m


43

Solution

With ( )0; mα∈ , using inequality AM – GM applied to two positive numbers, we have:

( ) ( ) ( )

22

22

2 2

22 2

22 2

2

ca ac

cb bc

m a b m ab

� αα + ≥

+ α α + ≥

− α + ≥ − α�

� ( ) ( ) ( )2 2 2 2 22

m a b c ac bc m abα+ + ≥ + + − α

Choose ( )0; mα∈ such that 2

mα = − α ⇔ 1 1 8

2 4

m− + +α = . Then we have:

( ) ( )2 2 2 1 1 8 1 1 8

2 2

m mS m a b c ab bc ca

− + + − + += + + ≥ + + =

Equality occurs ⇔ 4 4

1 1 81 ,1 8 2 1 8

ma b c

m m

− + += = =

+ ⋅ +. Thus

1 1 8Min

2

mS

− + +=

Problem 6. Let , , , , 0a b c m n > be satisfying 1ab bc ca+ + = .

Find the minimum value of 2 2 2S ma nb c= + + according to m, n

Solution

Supposing , , 0x y z > such that , ,1 0m x n y z− − − > . Using AM – GM inequality we have:

( ) ( )

( ) ( ) ( ) ( )

2 2

2 2

2 2

. . 2

. 2

1 2 1

x a y b xyab

m x a z c m x z ac

n y b z c n y z bc

� + ≥

+ − + ≥ −

− + − ≥ − −�

� ( ) ( ) ( )2 2 2 2 2 2 1S ma nb c xy ab m x z ac n y z bc= + + ≥ + − + − −

Equality occurs ⇔ ( )

( ) ( )( ) ( ) ( )

2 22 2

2 22 2

2 211

xb axa yb y

m xm x a zc c az

n y b z cn y xz z m x y

� =� =

⇔ −− = =

− = −� − = − −�

Choose , ,x y z such that ( ) ( ) ( )1 0xy m x z n y z k= − = − − = > � ( ) ( ) ( ) 31n y xz z m x y k− = − − =

We have: ( )[ ] ( ) ( )[ ] ( )2 31 1 2mn x m x y n y z z k m n k� = + − + − + − = + + +� �

Letting ƒ(k) = ( )3 22 1k k m n mn+ + + − � ƒ′(k) = ( ) ( )26 2 1 0, 0f k k m n k k′ = + + + > ∀ >

� ƒ(k) is increasing on ( )0; +∞ � The equation ƒ(k) = 0 has unique positive root 0

0k >

� ( )2 2 202 2S ma nb c k ab bc ca k= + + ≥ + + = �

0Min 2S k=


44

Problem 7. Let be given , ,a b c satisfy 2 2 2 1a b c+ + = . Find the maximum value of

( ) ( ) ( ) ( )P a b a c b c a b c= − − − + + (IMO 2006)

Solution

We have: ( ) ( ) ( ) ( ) ( )22 2 22 2 23 2 2a b c a b a c b c a b c� � + + = − + − − + + +� ��

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 28 2 16 2 16 2a c b c a b a b c a c b c a b a b c P� ≥ − − − + + + ≥ − − − + + ≥� �

� 9

16 2P ≤ . Equality occurs ⇔

3 3 6 6 6 3 3; ;

6 2 6 2 6 2a b c

+ −= = = . Thus 9Max

16 2P = .

Problem 8. (Phan Thanh Viet) Let 1 2, , , 0na a a ≥� be satisfying 1 2 1na a a+ + + =� .

Prove that S = ( ) ( ) ( ) 11 2 1 2 3 1 2 1 1 2..... ... 4n

n na a a a a a a a a a a−

−+ + + + + + ≥ �

Proof

With 1 2, , , 0nx x x ≥� and 1 2

... , 1,k k

S x x x k n= + + + ∀ = , using AM – GM inequality we have:

( )1 2

1 2 1 21 2 1 2 1

1 2 1 2

...

k

kk k

xx x

SS Sk k

k k kk k

a aa a a aa a a x x x x x

x x x x x x

� �� + + + = ⋅ + ⋅ + + ⋅ ≥ + + � ��

� � � � � ��

It follows: ( ) ( ) ( ) ( ) 1 22

1 2 1 2 3 1 1 1 1 2... ... . nCC C

n n nS a a a a a a a a a C a a a

−= + + + + + + + ≥� � �

1 1 1

12 1

2...

n n

x x xC

S S S−

= + + + ;1

2... , 2, 1k k k

kk n n

x x xC k n

S S S−

= + + + ∀ = − ; 2

n

nn

xC

S= ;

1 2

2

2 3 1

1 2

...

n

n n

CC C

n

S S S SC

x x x

−=�

Choose 1 1 1

1 1; 2,2 2

kn n kx x k n

− + −= = ∀ = so

1 1 2 1

1 1 1 1 1

2 2 2 2 2k n n n n k n k

S− − − + − −

= + + + + =�

� ( ) ( )2 3 1

1 1 1

1 12 2 ... 2 1 1; 2 2 .... 2 1 1, 2, 12 2

n n n k n k

kn n kC C k n

− − − − +

− + −= + + + + = = + + + + = ∀ = −

( ) ( )

( )

1 2 .... 2 2 11

1 2 ... 2

1 22 1 1 ; 42 2

n nn

n nC C

+ + + − + −−

+ + + −= ⋅ ⋅ = = = . Thus

11 24n

nS a a a−≥ � (q.e.d)

Equality occurs ⇔ 1 1 1

1 1; , 2,2 2

kn n ka a k n

− + −= = ∀ =

Problem 9. Letting that 1 2, , , nx x x� are n real numbers satisfying 1 2 0nx x x+ + + =� and

1 2 1nx x x+ + + =� . Find the maximum value of 1

i j

i j n

P x x≤ ≤ ≤

= −∏

Solution

Case 1: For 2n = , the condition becomes 1 2 1 20; 1x x x x+ = + =

� 1 2 1 2

1 1 1 1; or ;2 2 2 2

x x x x= = − = − = � 1 2

1P x x= − =


45

Case 2: For 3n = , WLOG supposing 1 2 3x x x≤ ≤ . We have ( ) ( )3 1

2 1 3 22 2

x xP x x x x−

= − − .

Using AM – GM we have: ( ) ( )

3

3 1 32 1 3 2

3 12 1 12 3 2 8 4

x xx x x x x xP P

−� �− + + −� � −� �

≤ = ≤ ⇔ ≤� ��

Equality occurs ⇔

1 2 3

1 2 3

3 1

2 1 3 2

0

1

2

x x x

x x x

x xx x x x

+ + =� + + = − − = = −�

⇔

1

2

3

12

0

12

x

x

x

� = −

=

=�

. Thus 1max4

P =

Case 3: For 4n = , WLOG supposing 1 2 3 4x x x x≤ ≤ ≤ , then we can predict Max P can be

obtained when 1 4 2 3;x x x x= − = − � 2 1 4 3x x x x− = − . Letting ( )3 2 4 3x x a x x− = − , P reaches

Max when the variables are satisfied with the conditions:

3 2 3 1 4 2 4 12 1 4 3 1 1 2

x x x x x x x xx x x x

a a a a

− − − −− = − = = = =

+ + +.

From these, we come to this solution: WLOG supposing 1 2 3 4x x x x≤ ≤ ≤

� ( ) ( ) ( ) ( ) ( ) ( )2 1 3 2 4 1 3 2 4 2 4 3P x x x x x x x x x x x x= − − − − − −

Hence ( ) ( )

( )( ) ( ) ( ) ( )

( )3 2 3 24 1 4 2

2 1 4 32 1 2 12 1

x x x xx x x xP x x x xa a a aa a a

− −− −= − −

+ + ++ +

( )( ) ( ) ( ) ( )

( )

6

3 2 3 24 1 4 2

2 1 4 3

16 1 2 1

x x x xx x x xx x x x

a a a a

� � �− −− −≤ − + + + + −� ��

+ + +� ��

It follows: ( ) ( ) ( ) ( )6

2 1 3 28

1 1 1 1 11 11 2 12

P x x x xa a a a

� ≤ − + + + + + − −� �+ + +� �

Choose a such that ( ) ( )1 1 1 11 11 2 1a a a a

+ + = + + −+ + +

⇔ 2 1a = − .

Now ( ) ( ) 3 21 1 1 11 11 2 1 2a a a a

+ + = + + − =+ + +

� ( ) ( ) ( )2 6

6

1 2 3 48

2 3 212 1 2 12 22

P x x x x� � �

+ − ≤ − − + +� � � ��

6

1

2P ≤

Equality occurs

⇔

( ) ( )

1 2 3 4

1 2 3 4 1 2 3 4

4 3 3 1 4 2 4 1

1 2 4 3

0

1

2 1 2 2 2 1

x x x x

x x x x x x x x

x x x x x x x xx x x x

+ + + =�− − + + = + + + = − − − − − = − = = = = − +�

⇔ 4 1

3 2

2 2

4

2

4

x x

x x

� −= − =

= − =�

(1)

Conclusion: 8

1Max2

P = (occurs when variables are satisfying the condition (1))


46

Case 4: 5n = : With such assumption 1 2 3 4 5x x x x x≤ ≤ ≤ ≤ , referring to the solutions to case 1

and 2, we can predict P reaches Max when 5 1 4 2 3, , 0x x x x x= − = − =

� 2 1 5 4 3 2 4 3,x x x x x x x x− = − − = − .

Letting ( )3 2 2 1x x a x x− = − , then P reaches Max when variables are satisfying the condition:

4 3 3 2 3 1 5 3 5 2 5 1 5 42 1 4 2 4 1

1 1 1 2 2 1 2 1 2 2 1

x x x x x x x x x x x x x xx x x x x x

a a a a a a a a

− − − − − − −− − −= = = = = = = = =

+ + + + +

From such analysis, the solution to the problem in which 5n = will be as follows:

WLOG supposing 1 2 3 4 5x x x x x≤ ≤ ≤ ≤ . Now we have:

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 1 3 1 4 1 5 1 3 2 4 2 5 2 5 2 5 4P x x x x x x x x x x x x x x x x x x= − − − − − − − − −

Consider the expression: ( ) ( )3 224 1 2 1

PQa a a

=+ +

. Transform Q in the form:

3 1 5 1 3 2 5 2 4 3 5 3 5 42 1 4 1 4 2

1 1 2 1 2 2 2 2 1 1 1

x x x x x x x x x x x x x xx x x x x xQ

a a a a a a a a

− − − − − − −− − −= ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅

+ + + + +

Using AM – GM applied to 10 non-negative numbers, we have:

3 1 5 1 3 22 1 4 1

10

10

5 2 4 3 5 3 5 44 2

11 1 2 1 2 210

2 2 1 1 1

x x x x x xx x x xQ

a a a a

x x x x x x x xx x

a a a a

− − −− −�≤ + + + + +� + + +�

− − − −− + + + + + �+ + �

It follows: ( )

( ) ( ) ( )10

1 5 2 40

3 31 1 11 12 1 2 2 12 110

Q x x x xa a aa1

� � �≤ + + − + + − + + − +� �� + ++� �� (2)

Choose 0a > such that ( ) ( )3 31 11 1

2 1 2 2 12 1q

a a aa

� �+ + = − + + =� �+ ++� � ⇔ 51 ,

2 2a q= =

It follows 27 4

PQ = . Now: ( )10

1 2 4 510 10

51 1210 2

Q x x x x� ≤ − − + + =� ��

⇔ 20 22

27 27142 2

P ≤ =

Equality occurs ⇔

( )

3

1 2 3 4 5

1 2 3 4 5 1 2 3 4 5

3 1 5 1 3 2 5 2 4 3 5 3 5 44 1 4 22 1

0

0

1

3 2 2 3 1 2 1 2 1 2 3 2 1

x

x x x x x

x x x x x x x x x x

x x x x x x x x x x x x x xx x x xx x

=�

+ + + + =− − + + + = + + + + =

− − − − − − −− − − = = = = = = = = =�

⇔ 1 5 2 4 33 1; ; 08 8

x x x x x= − = − = − = − = (3)

Conclusion: 22

27Max2

P = (occurs when variables are satisfying the condition (3))

Remarks. In general cases ( 6n ≥ ) the solution of the problem is still an open question


47

7. APPLYING AM – GM TO INEQUALITY OF DIFFERENT DEGREE

The un-homogeneous inequalities considering in positive real set:

Problem 1. (Tran Phuong) Let be given a, b, c > 0. Prove that

7 7 7

2 2 2 2 2 2 2 2 2 2 2 2

1bc ca ab a b cabc

a b c b c c a a b a b c+ + + ≤ + + +

Proof

Transfer and using AM – GM Inequality we have

7 7 7 7

32 2 2 2 2 2 2 2 2 2 2 2 2 2 2

7 7 7 7

32 2 2 2 2 2 2 2 2 2 2 2 2 2 2

7 7 7 7

32 2 2 2 2 2 2 2 2 2 2 2 2 2 2

7 7 7

32 2 2 2 2 2

1 1 1

3

1 1 1

3

1 1 1

3

1

3

bc b c b c

a a c a b a b c a c a b a b c

ca c a c a

b b a b c a b c b a b c a b c

ab a b a b

c c b c a a b c c b c a a b c

a b cabc

c b c a a b

� �= ⋅ ⋅ ≤ + +� �

� �

� �= ⋅ ⋅ ≤ + +� �

� �+

� �= ⋅ ⋅ ≤ + +� �

� �

= ⋅ ⋅ ≤7 7 7

2 2 2 2 2 2

a b c

c b c a a b

�

� � + +� � � ��

� 7 7 7

2 2 2 2 2 2 2 2 2 2 2 2

1bc ca ab a b cabc

a b c b c c a a b a b c+ + + ≤ + + +


9 9 95 5 52 2a b c a b c

bc ca ab abc+ + + ≥ + + + (1)

Proof

Using AM – GM Inequality we have: 9 9 9

5 5 52 ; 2 ; 2a b cabc a abc b abc cbc ca ab

+ ≥ + ≥ + ≥

� ( )9 9 9

35 5 5 5 5 5 5 5 52 3 3 3a b c a b c abc a b c a b c abcbc ca ab

+ + ≥ + + − ≥ + + + ⋅ −

Therefore, to prove inequality (1), we have to prove:

3 5 5 5 5 3

3

2 23 3 2 3 3 2a b c abc t tabc t

⋅ − + ≥ ⇔ − + ≥ (for 3 0t abc= > )

( ) ( )2 6 5 4 3 2

3

1 3 6 6 6 6 4 20

t t t t t t t

t

− + + + + + +⇔ ≥ (is always true) � (q.e.d.)

Equality occurs ⇔ 1a b c= = =


9 9 94 4 43 3a b c a b c

bc ca ab abc+ + + ≥ + + + (1)

Proof


48


9 9 92 4 2 4 2 43 , 3 , 3a b cabc a a abc b b abc c c

bc ca ab+ + ≥ + + ≥ + + ≥

� ( ) ( )9 9 9

4 4 4 2 2 23 3a b c a b c a b c abcbc ca ab

+ + ≥ + + − + + − (2)

Using again AM – GM inequality, we have:

( ) ( ) ( ) ( )4 2 4 2 4 2 4 4 4 2 2 231 1 1 11 ; 1 ; 12 2 2 2 2

a a b b c c a b c a b c+ ≥ + ≥ + ≥ � + + + ≥ + +

From (2) � ( )9 9 9

34 4 4 4 4 4 4 4 45 3 9 33 3

2 2 2 2

a b ca b c abc a b c a b c abc

bc ca ab+ + ≥ + + − − ≥ + + + ⋅ − −

Therefore, to prove inequality (1), we only have to prove:

3 4 4 49 3 33 32 2

a b c abcabc

⋅ − − + ≥ ⇔ 4 3

3

9 3 932 2

t tt

⋅ − + ≥ (for 3 0t abc= > )

⇔ ( ) ( ) ( )2 4 3 23 1 1 3 4 2 2 02

t t t t t t− + + + + + ≥ (is always true) � (q.e.d.)


Problem 4. (Tran Nam Dung) Let be given a, b, c ≥ 0. Prove that

( ) ( )2 2 22 8 5a b c abc a b c+ + + + ≥ + + (1)

Solution

� Lemma: (Schur inequality)

( ) ( ) ( )3 3 3 3x y z xyz xy x y xy y z zx z x+ + + ≥ + + + + + , , , 0x y z∀ ≥ (*)

� Proof: WLOG supposing 0x y z≥ ≥ ≥ . We have:

LHS (*) – RHS (*) = ( ) ( ) ( ) ( ) ( )2 2

0x x y z y z z x y x y y z− + − + + − − − ≥

� Application: Using AM – GM inequality and Schur inequality we have:

6.[LHS (1) – RHS (1)] ( ) ( )2 2 212 6 48 30a b c abc a b c= + + + + − + +

( ) ( ) ( )2 2 212 3 2 1 45 5 2 3a b c abc a b c= + + + + + − ⋅ ⋅ + +

( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

2 232 2 2

2 2 22 2 2

3

12 9 45 5 9

9 3 6 2

a b c abc a b c

abc a b c ab bc ca a b b c c aabc

� ≥ + + + ⋅ + − + + +� �

� = + + + − + + + − + − + −� �

( ) ( ) ( ) ( )22 2 2

3

9 273 6 3 12abc abca b c ab bc ca a b c ab bc caa b cabc

≥ + + + − + + ≥ + + + − + ++ +


49

( ) ( ) ( )33 9 4abc a b c ab bc ca a b ca b c

� = + + + − + + + +� �+ +

( ) ( ) ( )[ ]3 3 33 3 0a b c abc ab a b bc b c ca c aa b c

= + + + − + − + − + ≥+ +

� (q.e.d.)


Problem 5. (Darij Grinberg) Let be given a, b, c > 0. Prove that

( )2 2 2 2 1 2a b c abc ab bc ca+ + + + ≥ + +

Proof

Using AM – GM Inequality and Schur inequality we have:

( ) ( )2 2 2

2 2 2 2 2 2 3 3 32 1 2 3 2a b c abc ab bc ca a b c a b c ab bc ca+ + + + − + + ≥ + + + − + +

( ) ( ) ( ) ( )

( ) ( ) ( )

2 2 2 2 2 2 2 2 2 2 2 23 3 3 3 3 3 3 3 3 3 3 3

2 2 22 2 1 1 2 2 1 1 2 2 1 13 3 3 3 3 3 3 3 3 3 3 3

2

0

a b a b b c b c c a c a ab bc ca

a b a b b c b c c a c a

≥ + + + + + − + +

= − + − + − ≥


Problem 6. (APMO 2004) Let be given a, b, c > 0. Prove that

( ) ( ) ( ) ( )2 2 22 2 2 9a b c ab bc ca+ + + ≥ + + (1)

Proof


LHS (1) – RHS (1) = ( ) ( ) ( ) ( )2 2 22 2 2 9a b c ab bc ca+ + + − + +

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 2 2 2 24 2 1 1 1 1 1 9a b c a b b c c a a b c ab bc ca� = + + + + + + + + + + + − + +� �

( ) ( ) ( )2 2 24 4 2 1 9a b c ab bc ca abc ab bc ca≥ + + + + + + + − + +

( )2 2 2 2 1 2a b c abc ab bc ca≥ + + + + − + +

Using the results of problem 5 we have (q.e.d.). Equality occurs ⇔ 1a b c= = =

Problem 7. Let be given a, b, c, d > 0. Prove that

( ) ( ) { }1 1 1 1 1 1 1 1 1 116 Max ;1a b c d a b b c c d d a abcd ac bd

+ + + + + + ≥+ + + + + +

(1)

Solution


( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1 11 a b c d a b a b c d c dLHSab cd ab c d cd a b ab d a ab b c cd b c cd d a

+ + + + + +� = + + + + + + +� �+ + + + + +� �


50

( ) ( ) ( ) ( )4 a b a b c d c d

ab d a ab b c cd b c cd d aabcd

+ + + +≥ + + + ++ + + +

. Similarly we have:

( )( ) ( ) ( ) ( )

41 b c b c a d a dLHSbc a b bc c d ad a b ad c dabcd

+ + + +≥ + + + ++ + + +

. It follows:

( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

82 1 a b a d a b b cLHSab d a ad a b ab b c bc a babcd

c d b c c d a d

cd b c bc c d cd d a ad c d

+ + + +� � ⋅ ≥ + + + + +� � � �+ + + +� � � �

+ + + +� � + + + +� � � �+ + + +� � � �

( ) ( )8 82 2 2 2 2 1 1 2 1 1a c b dabcd a bd b ac c bd d ac abcd bd ac

≥ + + + + = + + + +

8 164 4

abcd abcd abcd abcd≥ + + = � ( ) 81LHS

abcd≥

On the other hand, using AM – GM Inequality we have:

{ }Min 1 ;1 ;2 2 2

abcd ac bdabcd ac bdabcd abcd abcd+ ++ +≥ ≥ � ≤

� ( ){ } { }16 1 11 16 Max ;

1Min 1 ;LHS

abcd ac bdabcd ac bd≥ =

+ ++ + (q.e.d.)

Problem 8. (Le Trung Kien)

Prove that ( ) ( )3 3 3 4 9 8a b c a b c abc ab bc ca+ + + + + + ≥ + + , , , 0a b c∀ ≥ (1)

Solution

Lemma: ( ) ( ) ( ) ( )4 4 4 2 2 2 2 2 2a b c abc a b c ab a b bc b c ca c a+ + + + + ≥ + + + + + (1)

Proof: WLOG supposing a b c≥ ≥ . We have:

(1) ⇔ ( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 2 2 2 2 2 2 2 0b c a b c c a b c a a b c a b+ − − + + − − + + − − ≥

( ) ( ) ( ) ( ) ( )2 22 2 2 2 2 0c b c a a b c a b a b b c⇔ − + − + + − − − ≥ (is always true)

Application: Using AM – GM inequality we have:

( )( )

( )2

44 8

ab bc caa b c ab bc ca

a b c

+ ++ + + ≥ + +

+ +

So it suffices to prove that: ( ) 2

3 3 3 49

ab bc caa b c abc

a b c

+ ++ + + ≥

+ +

⇔ ( ) ( ) ( ) ( ) ( )4 4 4 2 2 2 2 2 2 2 2 2 2 2 24a b c abc a b c ab a b bc b c ca c a a b b c c a+ + + + + + + + + + + ≥ + + (*)

Using lemma we have:

LHS (*) ( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 2 2 2 22 4ab a b bc b c ca c a a b b c c a� ≥ + + + + + ≥ + +� � � (q.e.d)

Equality occurs ⇔ 1a b c= = = or ( ) ( ), , ~ 2; 2;0a b c


51

8. SPECIALIZATION IN INEQUALITY OF DIFFERENT DEGREE


3

a b c

a b c abc

>�


3 3 3

1 1 1 3a b c

+ + ≥ (1)

Proof

1 1 13 3a b c abcab bc ca

+ + = � + + = . Using AM – GM Inequality we have:

( )3 3 3 3 3 3 3 3 3

1 1 1 1 1 1 1 1 1 1 1 12 3 1 1 1 3 9ab bc caa b c a b b c c a

� � � � � � � �+ + + = + + + + + + + + ≥ + + =� � � � � � � ��

� (q.e.d)

Problem 2. Given

, , 0

3

a b c

b c aa b cc a b

>�

+ + =�

. Prove that 6 6 6

3 3 33a b c

b c a+ + ≥ (1)

Proof


6 6 6 6 6 6

3 3 3 3 3 34 4 4 6a b b c c a b c aa b c

c a bb c c a a b

� �� + + + + + + + + ≥ + +� ��

� ��

⇔ 6 6 6 6 6 6

3 3 3 3 3 32 12 18 3a b c a b c

b c a b c a

� �+ + + ≥ ⇔ + + ≥� �

� � (q.e.d)

Problem 3. Given , , , 0

1

a b c d

ab bc cd da

>�

+ + + =�. Prove that

3 3 3 313

a b c d

b c d c d a d a b a b c+ + + ≥

+ + + + + + + + (1)


Proof

Using AM – GM inequality we have:

( ) ( )

( ) ( )

( ) ( )

( ) ( )

3 34

3 34

3 34

3 34

36 362 6 3 4 2 6 3 24

36 362 6 3 4 2 6 3 24

36 362 6 3 4 2 6 3 24

36 362 6 3 4 2 6 3 24

a ab c d a b c d a ab c d b c d

b bc d a b c d a b b

c d a c d a

c cd a b c d a b c c

d a b d a b

d da b c d a b c da b c a b c

+ + + + + ≥ ⋅ ⋅ + + ⋅ ⋅ =+ + + +

+ + + + + ≥ ⋅ ⋅ + + ⋅ ⋅ =+ + + +

+

+ + + + + ≥ ⋅ ⋅ + + ⋅ ⋅ =+ + + +

+ + + + + ≥ ⋅ ⋅ + + ⋅ ⋅ =+ + + +

d

��

� 3 3 3 3

13 3

a b c d a b c d


+ + ++ + + ≥ −+ + + + + + + +

On the other hand, ( ) ( )24 4 2a b c d ab bc cd da a b c d+ + + ≥ + + + = � + + + ≥

� LHS (1) ≥ 1 2 1 13 3 3 3 3

a b c d+ + + − ≥ − = (q.e.d)


52

9. BEST SOLUTIONS TO FOUR TRIGONOMETRIC INEQUALITIES

Let ABC be a triangle. Prove that

T1 = 3

sin sin sin2 2 2 2

A B C+ + ≤ ; T2 = 3cos cos cos

2A B C+ + ≤

T3 = 3 3

sin sin sin2

A B C+ + ≤ ; T4 = �

��

�

��

�

�� ++ ≤

�

��

Proof

T1 = sin sin sin sin sin .1 cos cos sin sin2 2 2 2 2 2 2 2 2

A B C A B A B A B� �+ + = + + −� �

� �

≤

2

2 2 21 1 3sin sin 1 cos cos sin sin

2 2 2 2 2 2 2 2 2

A B A B A B� � � � �+ + + + − =� ��

� � � ��

T2 = ( )cos cos cos cos cos .1 sin sin cos cosA B C A B A B A B+ + = + + −

≤ ( )2 2 2 21 1 3(cos cos ) 1 sin sin cos cos

2 2 2A B A B A B� + + + + − ≤� �

T3 = 2 3 3 sin sin

sin sin 3 cos cos2 23 3 3

A BA B B A

� � � �⋅ + ⋅ + ⋅ + ⋅� � � ��

≤ ( ) ( )2 2

2 2 2 21 3 3 33 3 sin sinsin sin cos cos4 4 3 32 23

A BA B B A

� � � � �� + + + + + + + =� � � ��

T4 = cos cos

2 3 3 2 2cos cos 3 sin sin2 2 2 2 2 23 3 3

A BA B B A

� ��

⋅ + ⋅ + ⋅ + ⋅� � � ��

≤

2 2

2 2 2 2cos cos

1 3 3 3 3 32 2cos cos sin sin2 4 2 4 2 3 2 3 2 23

A BA B B A

� � � � ��

+ + + + + + + =� � � � � � � ��

Comment: In my colleagues’ opinions, the common solutions to the four above problems are

the most unprecedented and the shortest among what have ever demonstrated in the world. In

all other publications, these inequalities are proved based on graphs of Jensen Inequality but

the above solutions originally apply a simple inequality 2 22xy x y≤ + and the concept "point

of incidence", which will be referred to in detail in the book. These solutions were published

in the other books in Viet Nam, also written by the author in 2000.

Trivial classical method:

� ( )sin sin 2sin cos 2sin , , 0,2 2 2

x y x y x yx y x y

+ − ++ = ≤ ∀ ∈ π

� sin sin sin sin 2sin cos 2sin cos 2sin 2sin2 2 2 2 2 2

x y x y x yz t z t z tx y z t

+ − ++ − ++ + + = + ≤ +

4sin cos 4sin4 4 4

x y z t x y z t x y z t+ + + + − − + + += ≤ , ( ), , , 0;x y z t∀ ∈ π

� Take , , ,3

A B Cx A y B z C t + += = = = � 3 3

sin sin sin 3sin 3sin3 3 2

A B CA B C + + π+ + ≤ = =


53

10. SELECTIVE PROBLEMS IN USING POINT OF INCIDENCE FOR AM – GM

Problem 1. (Tran Phuong) Let be given ∆ABC with the lengths of the sides a, b, c.

Prove that:

2 2 2

33 3 33 3 3 3

453 3 3 3 3 3 5

a b c

b c a c a b a b c

� � � � � �+ + ≥ ⋅� � � � � �

+ − + − + −� � � � � � (1)

Proof

Let

3 1 3 3 4

3 3 3 3

3 1 3 3 4

3 3 3 3

3 3 43 1

3 33 3

a b c a a b cx x

b c a x a a

b c a b a b cy y

c a b y b b

a b c a b cc x zc ca b c z

� + − + +� �= = + = + −

+ − + + = � = � + =

+ −

+ − + + = + == � �+ −�

� 1 1 1 14 4 43 3 3

a b c

a b cx y z

+ ++ + = =

+ ++ + +

1 1 14 4 4

13 3 3

x y z

− − −� � � � � �

⇔ + + + + + =� � � � � ��

(1) ⇔ 3

33 2 2 23

1 1 1 345

5x y z+ + ≥ ⋅ .

Using AM GM− Inequality for 9 terms we have:

( )

( )

( )

9 9

4 9

5

9 9

4 9

5

9 9

4 9

5

4 455 53 3

3 9 3 3

4 455 53 3

3 9 3 3

4 455 53 3

3 9 3 3

x x

x

y y

y

z z

z

� � � �+ + � �� ⋅ ≤ = ⋅ � ��

� � � + +� �� ⋅ ≤ = ⋅ � ��

� � � + +� �� ⋅ ≤ = ⋅ � ��

9

5

5

9

5

5

9

5

5

45 3

3 3

45 3

3 3

45 3

3 3

x

x

y

y

z

z

� � �+ � � � �≤ ⋅ � ��

� �� +

⇔ � � � � ≤ ⋅ � ��

� �+� � � � ≤ ⋅ � ��

� ��

699 5

55

3

3 2

9955

3

23

9 95 5

444 1 9 31 35 33

25 35 33 3

4441 9 31 35 33

25 35 33 3

4 45 1 33 33 3 35

xxx

xxx

yyy

yy y

z zz

z

−−

−

−

�� +� �� ++ � �� ≥ � � ≥ ⋅≤ ⋅ � ��

� � �� + ++⇔ ⇔ ⇔ � �� ≥≥ ⋅≤ ⋅ � ��

� � � � + +� � � � ≤ ⋅ ≥ ⋅� � � �

� � � ��

65

65

3

3 2

41 9 3

25 3

z

z

−

−

� �

� ��

� � +� � ≥ � � � ��

Letting ( ) ( ) ( )1 1 1

4 4 4; ; 13 3 3

m x n y p z m n p− − −

= + = + = + � + + =


54

Using AM GM− Inequality for 6 terms we have:

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

6 6 6 6 6 6

5 5 5 5 5 5

6 6 6 6 6 6

5 5 5 5 5 5

6 6 6 6 6 6

5 5 5 5 5 5

5. 3 1 3 3 3 3 3 1 18

5. 3 1 3 3 3 3 3 1 18

5. 3 1 3 3 3 3 3 1 18

m m m m m m m

n n n n n n n

p p p p p p p

�+ = + + + + + ≥

+ + = + + + + + ≥ + = + + + + + ≥�

( ) ( ) ( )( )6 6 6

33 35 5 533 2 2 23

18 31 1 1 9 9 33 3 3 45

25 25 5 5

m n pm n p

x y z

� + + −� � + + ≥ + + ≥ ⋅ =� ��



3 3 3 3 3 3 2 2 2 2 2 2

123 3 34 4 4

3 3 3 2 2 22


c a b c a b

� �+ + + + + +� �+ + ≥ + +� ��

Proof

� Lemma: 3 3 2 2

3

2 2

a b a b+ +≥ (It is for you to prove it)

� Application: Using AM – GM Inequality and the lemma we have

2

3 3 3 3 3 3 3 3 2 233

34 4 43 3 3 2 2

8 terms

3 3 3 3 3 3 3 3

4 4 43 3 3 2

8 terms

1... 1 9 9

22 2 2 2

1... 1 9

22 2 2

a b a b a b a b a b

c c c c c

b c b c b c b c

a a a a

� �+ + + + ++ + + + ≥ ⋅ ≥ ⋅� �

� �

�+ + + ++ + + + ≥ ⋅

��

��

2

2 233 3

2

2

3 33 3 3 3 3 3 2 233

34 4 43 3 3 2 2

8 terms

92

1... 1 9 9

22 2 2 2

b c

a

c ac a c a c a c a

b b b b b

� � +

≥ ⋅� ��

� �++ + + + + + + + ≥ ⋅ ≥ ⋅� � � ��

� 3 3 3 3 3 3 2 2 2 2 2 2

3 3 34 4 43 3 3 2 2 2

8 3 92 2 2 2 2 2


c a b c a b

� � � �+ + + + + +� � � �+ + + ≥ + +� � � ��

2 2 2 2 2 2 2 2 2 2 2 2

3 3 3 3 3 32 2 2 2 2 2

82 2 2 2 2 2


c a b c a b

� � � �+ + + + + +� � � �= + + + + +� � � ��

2 2 2 2 2 2 2 2 2 2 2 2

3 3 3 92 2 2 2 2 2

8 3.2 2 2 2 2 2


c a b c a b

� �+ + + + + +� �≥ + + + ⋅ ⋅� ��

2 2 2 2 2 2

3 3 3 92 2 2 2 2 2

2 2 28 3.

2 2 2 2 2 2

a b b c c a ab bc ca

c a b c a b

� �+ + +� �≥ + + + ⋅ ⋅� ��


55

2 2 2 2 2 2

3 3 32 2 2

8 32 2 2

a b b c c a

c a b

� �+ + +� �= + + +� ��

. It follows

3 3 3 3 3 3 2 2 2 2 2 2

3 3 34 4 43 3 3 2 2 22 2 2 2 2 2


c a b c a b

+ + + + + ++ + ≥ + + � (q.e.d.)


Problem 3. (Vo Quoc Ba Can) Let be given a, b, c > 0. Prove that

32 2 2

3 3 32 2 2 2 2 2

9 abca bc b ca c ab

a b cb c c a a b

⋅+ + ++ + ≥+ ++ + +

(1)

Proof

Using AM GM− Inequality we have:

( ) ( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 2 23

2 2 23

a b c b c a c a b a b c abc b cb c

a bc a bc a bc

+ + + + + + += + + ≥ ⋅

+ + +

( )( ) ( ) ( )

2 323

2 2 2 2 2 2 2 2

3 a bc abca bc

b c a b c b c a c a b

+ ⋅+� ≥+ + + + + +

( )( )( ) ( ) ( )

2 2 2 32

32 2 2 2 2 2 2 2

31

cyc

a b c ab bc ca abca bcLHSb c a b c b c a c a b

+ + + + + ⋅+� = ≥+ + + + + +

� (2)

Using Schur Inequality we have:

( ) ( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 23a b c a b c ab bc ca a b c b c a c a b� + + + + + + + − + + + + +� �

( ) ( ) ( )3 3 3 3 0a b c abc ab a b bc b c ca c a= + + + − + − + − + ≥

� ( )

( ) ( ) ( )

2 2 2

2 2 2 2 2 2

3 9a b c ab bc ca

a b ca b c b c a c a b

+ + + + +≥

+ ++ + + + + (3)

From (2) and (3) follow ( )39

1abc

LHSa b c

⋅≥

+ + (q.e.d.)

Equality occurs ⇔ 0a b c= = ≥

Problem 4. Let be given a, b, c > 0. Prove that

( ) ( ) ( ) ( )2 2 2

1 1 1a b c b c a c a ba b c

a bc b ca c ab a b c

+ + + � �+ + ≤ + + + +� �+ + + � � (1)

Proof

(1) ⇔ ( ) ( )

2

2

1 1 1

cyc

a b ca b c

a bc a b c

� �+ � �≤ + + + +� � � �� + � ��

( ) ( ) ( )

( ) ( )2 2 22 3

cyc cyc cyc

a b c ab a c b c b c

a bc bca bc b ca

+ + + +⇔ + ≤ ++ + +

� � �


56

( ) ( ) ( )

( ) ( )2 2 22 3 0

cyc cyc cyc

a b c ab a c b c b c

a bc bca bc b ca

+ + + +⇔ + − − ≤+ + +

� � � (2)

Using AM GM− Inequality we have: ( ) ( )2 2 22cyc cyc cyc

a b c a b c b c

a bc bca bc

+ + +≤ =+

� � �

On the other hand, ( ) ( ) ( ) ( ) ( ) ( )22 2 0a bc b ca ab a c b c c a b a b+ + − + + = − + ≥

( ) ( )

( ) ( )2 21

ab a c b c

a bc b ca

+ +� ≤

+ +

( ) ( )

( ) ( )2 23

cyc

ab a c b c

a bc b ca

+ +� ≤

+ +�

( )2 6 3 12 2cyc cyc cyc

b c b c b cLHSbc bc bc

+ + +� �� ≤ + − − = −� ��

� � �( )

2

02cyc

b c

bc

−= − ≤�

� (2) is true � (q.e.d.). Equality occurs ⇔ 0a b c= = >

Problem 5. Let be given a, b, c, d > 0. Prove that

( )3 3 3 3 3 3 3 3 3 3 3 3 3

2431 1 1 1 1 1

2a b a c a d b c b d c d a b c d+ + + + + ≥

+ + + + + + + + + (1)

Proof

WLOG supposing 0a b c d≥ ≥ ≥ ≥ � ( ) ( )3 3

3 3

3 3d da b a b+ ≤ + + + ; ( )

33 3

3da d a+ ≤ +

( ) ( )

( ) ( )

( ) ( )

3 3 3 3

3 3 3 3

3 3 3 3

1 1

3 3

1 1

3 3

1 1

3 3

a b d da b

a c d da c

a b d db c

� ≥ +

+ + + ≥

� ++ + +

≥ +

+ + +�

;

( )

( )

( )

3 3 3

3 3 3

3 3 3

1 1

3

1 1

3

1 1

3

a d da

b d db

c d dc

� ≥ +

+ ≥

++

≥ +

+�

� ( )3 3 3

1 11cyc cyc

LHSx y x

≥ ++

� � for , ,3 3 3d d dx a y b z c= + = + = +

We need to prove ( )

3 3 3 3

2431 1

2cyc cycx y x x y z+ ≥

+ + +� � ⇔

( )3 3 3 3 3

2432 1 1

cyc x y x y x y z

� �+ + ≥� �+ + +� ��

Using AM – GM Inequality we have ( )

33 3 3 3 3 3 3 3

2 1 1 23x y x y x y x y

+ + ≥ ⋅+ +

( ) ( ) ( )

( )( )

33 4 32 2 2 23

2 2 243 3

3

4

xy x xy y x y x yxy x xy yx y

= ⋅ ≥ ⋅ =− + + +� �+ − +

+� ��


57

� ( ) ( ) ( ) ( )3 3 3 3 3

32 1 1 124 24cyc cyc x y y z z xx y x y x y

� �+ + ≥ ≥ ⋅� � + + ++ +� ��

( ) ( ) ( ) ( ) ( )

3 3 3

72 27 24372.2

3

x y zx y y z z x x y z≥ = =

+ +� � + + + + + + +� ��

� (q.e.d.)

Equality occurs ⇔ ( ), , ,a b c d is a permutation of (1; 1; 1; 0)

Problem 6. Let be given a, b, c > 0 satisfying the condition 1a b c+ + =

Find the maximum value of abca bS

a bc b ca c ab= + +

+ + +

Solution

abca bS

a bc b ca c ab= + +

+ + +

1 1 /

1 1 1

ab c

bc ca ab

a b c

= + ++ + +

Taking 2 2tan ; tan2 2

bc A ca B

a b= = for 0 < A, B < π

Then 1 ab ca bc ab ca bca b cc b a c b a

= + + = ⋅ + ⋅ + ⋅

It follows ( )1 tan tan

2 2 cot tan2 2tan tan

2 2

A Bab CA B

A Bc

− ⋅+= = =

+ for

, , 0A B C

A B C

>�

+ + = π�

S = ( )2 2

22 2

tan1 1 sin 12 cos cos 1 cos cos sin

2 2 2 21 tan1 tan 1 tan22 2

CA B C

A B CA B C

+ + = + + = + + +++ +

( )3 31 11 cos cos 3 sin .cos 3 sin .cos2 23 2 3

A B A B B A� �

= + + + +� ��

( ) ( ) ( ) ( )2 2 2 2 2 23 31 11 cos cos 3sin cos 3sin cos4 42 3 4 3

A B A B B A� � ≤ + + + + + + + +� ��

( ) ( )2 2 2 23 3 3 3 31 cos sin cos sin 1

4 4 4 4A A B B= + + + + + = +

Equality occurs ⇔ 2

,6 3

A B Cπ π

= = =

tan 2 312

bc ca

a b

π⇔ = = = − , tan 3 2 3 3 , 7 4 33

ab a b cc

π= = ⇔ = = − = −

Inequality Proof by Analysis of Sum of Squares (S.O.S) 58

§4.18. INEQUALITY PROOF BY ANALYSIS OF SUM OF SQUARES (S.O.S)

I. Express non-negative polynomial in terms of sum of squares

1. Definition: Polynomial ( )1 2, ,..., nF a a a is an n-variable function ( ) ( )1 2, ,..., ,...,k na a a a a=

and ( )1 2! , ,..., nF a a a� is sum of !n expressions coming from ( )1 2, ,..., nF a a a by all permutations

of all variables ( ) ( )1 2, ,..., na a a a= .

� Polynomial ( )1 2, , ..., nF a a a is an m-degree homogeneous function

⇔ ( ) ( )1 2 1 2, ,..., . , ,...,mn nF ta ta ta t F a a a=

� Polynomial ( )1 2 1 2, , ..., 0 , ,..., Dn nF a a a a a a≥ ∀ ∈ ⇔ ( )1 2, , ..., nF a a a is called as non-negative in

domain D.

2. Identity Hurwitz – Muirhead and inequality AM −−−− GM (Cauchy)

Consider the special case: ( ) 1 21 2 1 2, , ..., .... n

n nF a a a a a aα α α= for 0, 0k ka ≥ α ≥

Let [ ] [ ] ( ) 1 21 2 1 2 1 2

1 1, , ... ! , , ..., ! ....! !

nn n nF a a a a a a

n nα α αα = α α α = =� �

Specially, [ ] ( ) ( ) ( )1 21 2

...1 !1,0,0,...0 ...!

nn

a a an a a a A an n

+ + +−= + + + = = : Arithmetic Mean (AM)

( )1 1 1

1 2 1 2!1 1 1, , ..., .... ....!

n n n nn n

n a a a a a a an n n n� � = = =� ��

� : Geometric Mean (GM)

Consider transformation: ( ) ( ) [ ] [ ]1 21 2

....... , 0, 0,...0 1,1,1,...,1

n n nnn n

n

a a aA a a a a a n

n+ + +

− = − = −�

[ ] [ ] [ ] [ ]

[ ] [ ] [ ] [ ] [ ] [ ]

( ) ( ) ( ) ( )1 1 2 21 2 1 2 1 2 1 2 3 1

, 0, 0, ...0 1,1, 0, ..., 0 1,1, 0,..., 0 2,1,1,0, ..., 0

2,1,1, 0, ..., 0 3,1,1,1, 0,..., 0 ... 3,1,1, ...,1, 0,0 2,1,1, ...,1, 0 2,1,1, ...,1, 0 1,1,1, ...,1

1 ! ! !2 !

n n n n n

n n n n

n n

a a a a a a a a a an

− − − − −

= − − + − − − +

+ − − − + + − + −

= − − + − − +� � ( ) ( )3 32 1 2 3 4 ...na a a a a−� �− − +� ��

As ( ) ( ) 0r rp q p qa a a a− − ≥ , 0p qa a∀ ≥ so ( ) ( ) 0n nA a a− ≥�

Let 2 21 1 ;...; n na x a x= = and ( )1 2, , ..., ni i i as a permutation of ( )1, 2,...n . Then we have:

( ) ( ) ( )1 2 1 2... ....n n n n nn nn A a a a a a na a a� �− = + + + − =� �� 2 2 2 2 2

1 1 2... ...n nn nx x nx x x+ + − =

( ) ( ) ( ) ( ) ( ) ( )1 2 1 2 1 2 1 2 3 1 2 3 4

22 2 2 2 2 2 2 4 2 4 2 2 2 2 2 2 2 21 ! ! ... ! ...2 1 ! n

n n n ni i i i i i i i i i i i i ix x x x x x x x x x x x x x

n− − − − ��= − − + − − + + − �� − � � �

Besides,

( ) ( ) ( )1 2 1 2 3 1 1 2 1 1 2 2 3 1

22 2 2 2 2 2 2 2 2 2 2 2 2 2 2 4 2 2 2 2 2 2... ... ...k k

n k n k n k n k n ki i i i i i i i i i i i i ix x x x x x x x x x x x x x

+ +− − − − − − − −� �− − = − + + +� �

( ) ( ) ( )1 2 3 1 1 2 3 11 2 1 2

222 1 12 2 2 2 2 2

1 1

... ...k k

n k n kn k j j n k j j

i i i i i i i ii i i ij j

x x x x x x x x x x x x+ +

− −− − − − − −

= =

� �� = − = −� � � ��

� �

� ( ) ( ) 2 2 21 2 ... 0n n

mn A a a s s s� �− = + + + ≥� �� ( ) ( )n nA a a≥ � � ( ) ( )A a a≥ �


3. Special cases: • n = 2: ( )22 21 2 1 2 1 22x x x x x x+ − = −

• n = 3: ( ) ( ) ( ) ( )2 226 6 6 2 2 2 2 2 2 2 2 2 2 2 21 2 3 1 2 3 1 2 3 1 2 2 3 3 1

132

x x x x x x x x x x x x x x x� �+ + − = + + − + − + −� �

• n = 4: ( ) ( ) ( )22 24 4 4 4 2 2 2 21 2 3 4 1 2 3 4 1 2 3 4 1 2 3 44 2x x x x x x x x x x x x x x x x+ + + − = − + − + −

• n = 5: 10 10 10 10 10 2 2 2 2 21 2 3 4 5 1 2 3 4 55x x x x x x x x x x+ + + + − =

( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 1 2 1 2 1 2 3 1 2 1 2 3 4 1 2 3 4 5

28 8 2 2 6 6 2 2 2 4 4 2 2 2 2 2 2 2 2 21 ! ! ! !2.4! i i i i i i i i i i i i i i i i i i i ix x x x x x x x x x x x x x x x x x x x� �= − − + − − + − − + −� ��

( ) ( ) ( ) ( )1 2 1 2 3 1 2 3 4 1 2 3 4 51 2 1 2 1 2

4 3 22 2 2 24 1 3 1 2 12 2 2 2 2 2 2 2

1 1 1

1 ! ! ! !2.4!

j j j j j ji i i i i i i i i i i i i ii i i i i i

j j j

x x x x x x x x x x x x x x x x x x x x− − − − − −

= = =

� �� = − + − + − + −� ��

��

• n = 6: ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

2 226 6 6 6 6 6 2 2 2 2 2 2 2 2 21 2 3 4 5 6 1 2 3 4 5 6 1 2 3 1 2 2 3 3 1

2 2 2 22 2 2 2 2 2 2 2 24 5 6 4 5 5 6 6 4 1 2 3 4 5 6

162

1 32

x x x x x x x x x x x x x x x x x x x x x

x x x x x x x x x x x x x x x

� �+ + + + + − = + + − + − + − +� �

� �+ + + − + − + − + −� �

4. Theorem and examples: 4.1. Theorem Hilbert 1:

If polynomial ( )1 2, , ..., nF a a a is homogeneous and non-negative in domain D, i.e.:

( )1 2 1 2, , ..., 0 , ,..., Dn nF a a a a a a≥ ∀ ∈ then it is possible to express

( ) 2 2 21 2 1 2, , ..., ...n mF a a a p p p= + + + in which kp is a certain real rational function.

4.2. Theorem Hilbert 2: If polynomial ( )1 2, , ..., nF a a a is non-homogeneous and non-negative

in domain 1 2 1 20, 0,... 0, ... 1n na a a a a a≥ ≥ ≥ + + + ≤ then

( ) ( ) 11 21 2 1 2 1 2, , ..., ... 1 ... nn

n k n nF a a a c a a a a a a +αα α α= − − − −�

in which 1 1, ... ,n n+α α α are non-negative integers and ck > 0.

4.3. Comment: The Theorems Hilbert 1 & 2 are qualitative, helping us believe that there

exists a way to prove a polynomial as non-negative by analysis of sum of squares (as for

homogeneous functions) or into sum of non-negative quantities (as for non-homogeneous

functions). It is, however, due to this obscure existence that it has little value in effect of

proof of elementary math inequalities. Take 3 extreme examples below to show that the

analysis into of squares would be actually difficult without belief in the existence of the

analysis. Examples 1&2 have short solutions but lack naturalness, so to find solutions would

be nothing more than words puzzle in which only a result is a meaningful word.

Example 1: ( )4 4 4 3 3 3 3 3 32a b c ab bc ca a b b c c a+ + + + + ≥ + + , ,a b c∀ ∈ ¡

Proof: ( )4 4 4 3 3 3 3 3 32 2a b c ab bc ca a b b c c a� �+ + + + + − + +� � =

= ( ) ( ) ( )2 2 22 2 2 2 2 2 0a ab ac c b bc ba a c ca cb b− + − + − + − + − + − ≥


Example 2: ( ) ( )22 2 2 3 3 33a b c a b b c c a+ + ≥ + + , ,a b c∀ ∈ ¡ (1)

1st proof:

(1) ⇔ ( ) ( )4 4 4 2 2 2 2 2 2 3 3 32 3a b c a b b c c a a b b c c a+ + + + + ≥ + + ⇔ 4 2 2 32 3cyc cyc cyc

a a b a b+ ≥� � �

⇔ S = 4 2 2 2 2 2 3 23 3 0cyc cyc cyc cyc cyc cyc

a a b a b a bc a b a bc − + − − − ≥� � � � � � � � ��

Transform: 4 2 2 2 2 21 ( )2cyc cyc cyc

a a b a b− = −� � � ; 2 2 2 213 ( 2 )2cyc cyc cyc

a b a bc ab ac bc − = + −� � ��

3 2 3 2 2 2

2 2 2 2 2 2

3 3 3 ( )

3 ( ) ( )( ) ( )( 2 )

cyc cyc cyc cyc cyc

cyc cyc cyc

a b a bc b c a bc bc a b

bc a b ab bc ca a b a b ab ac bc

− = − = − −� � � � � �

= − − + + + − = − + −

� � � � �

� � �

Then S = 2 2 2 2 2 21 1( ) ( 2 ) ( )( 2 )2 2cyc cyc cyc

a b ab ac bc a b ab ac bc− + + − − − + −� � �

= 2 2 21 ( 2 ) 02 cyc

a b ab ac bc− − − + ≥� � (q.e.d)

2nd proof: ( ) ( ) ( )22 2 2 2 2 2 3 3 34 3a b c ab bc ca a b c a b b c c a� �+ + − − − + + − + +� � =

( ) ( ) ( )

( ) ( ) ( )

23 3 3 2 2 2 2 2 2

23 3 3 2 2 2 2 2 2

5 4

3 2 6 0

a b c a b b c c a ab bc ca

a b c a b b c c a ab bc ca abc

� �= + + − + + + + + +� �

� �+ + + − + + − + + + ≥� �

II. METHOD OF ANALYZING SUM OF SQUARES (S.O.S.)

1. Introduction: To prove a homogeneous polynomial ( )1 2, , ..., 0nF a a a ≥ , according to

theorem Hilbert there exists an expression ( ) 2 2 21 2 1 2, , ..., ...n mF a a a p p p= + + + in which kp is a

certain real rational function, but to specify a certain expression would be very difficult. The

following method of analyzing sum of squares helps us surmount the abstraction of theorem

Hilbert, particularly to symmetrical functions ( ), ,F a b c . In Vietnam, this method was studied

and then stated by Tran Phuong, Tran Tuan Anh and Anh Cuong in August, 2004. In 2006,

Pham Kim Hung gave it the official name of S.O.S.

2. About S.O.S.

Let consider inequality A ≥ B in which A, B are expressions of variables , ,a b c .

Supposing we can turn the inequality A − B ≥ 0 into:

( ) ( ) ( )2 2 2 0a b cS A B S b c S c a S a b= − = − + − + − ≥ (1). Considers the following propositions:

2.1. Proposition 1: If 0 ; 0 ; 0a b cS S S≥ ≥ ≥ then ( ) ( ) ( )2 2 2 0a b cS S b c S c a S a b= − + − + − ≥

(1) is obviously true in this case. Therefore, we hereinafter consider other conditions.


2.2. Proposition 2: Letting that , , , , ,a b ca b c S S S are real numbers satisfying 2 conditions:

i) 0 ; 0 ; 0a b b c c aS S S S S S+ ≥ + ≥ + ≥ .

ii) If a b c≤ ≤ or a b c≥ ≥ then 0bS ≥

Then we have ( ) ( ) ( )2 2 2 0a b cS S b c S c a S a b= − + − + − ≥ (1)

Proof

WLOG suppose a b c≤ ≤ , then 0bS ≥ and

(1) ⇔ ( ) ( )[ ] ( ) ( )2 2 2 0b a cS c b b a S b c S a b− + − + − + − ≥

⇔ ( ) ( ) ( ) ( ) ( ) ( )2 2 2 0a b b c bS S b c S S a b S c b b a+ − + + − + − − ≥ (is always true)

2.3. Proposition 3: Letting that , , , , ,a b ca b c S S S are real numbers satisfying condition:

If a b c≤ ≤ or a b c≥ ≥ then 0, 0a cS S≥ ≥ and 2 0; 2 0a b c bS S S S+ ≥ + ≥


Proof WLOG suppose a b c≤ ≤ and 0, 0a cS S≥ ≥ ; 2 0 ; 2 0a b c bS S S S+ ≥ + ≥

If 0bS ≥ then (1) is always true.

If 0bS < then ( ) ( ) ( ) ( ) ( ) ( )2 2 2b a b c bS S S c b S S b a S c b b a= + − + + − + − −

( ) ( ) ( ) ( ) ( ) ( )2 2 2 2b a b c bS S c b S S b a S c b b a� �≥ + − + + − + − + −� �

( ) ( ) ( ) ( )2 22 2 0a b c bS S c b S S b a= + − + + − ≥

2.4. Proposition 4: Letting that , , , , ,a b ca b c S S S are real numbers satisfying condition:

If a b c≤ ≤ then 0, 0a bS S≥ ≥ and 2 2 0c bb S c S+ ≥


Proof

WLOG suppose a b c≤ ≤ and 0, 0a bS S≥ ≥ ; 2 2 0c bb S c S+ ≥

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )

2 22 2 2 2

22 2 2 0

a b c a b c

a b c

c a cS S b c b a S S S b c b a S Sb a b

b aS b c c S b Sb

� � � �−= − + − + ≥ − + − +� � � �−� � � �

−= − + + ≥

2.5. Proposition 5: Letting that , , , , ,a b ca b c S S S are real numbers satisfying 2 conditions:

i) 0 0 0a b b c c aS S S S S S+ > ∨ + > ∨ + >

ii) 0a b b c c aS S S S S S+ + ≥



Proof

WLOG supposing 0b cS S+ > . Letting ;u b a v c b= − = − , then:

S = ( ) ( ) ( )[ ] ( )22 2a b cS b c S c b b a S a b− + − + − + −

= ( ) ( ) ( ) ( ) ( ) ( )2 22b c b a bS S a b S c b b a S S b c+ − + − − + + −

( )2

2 0b a b b c c ab c

b c b c

S S S S S S SS S u v v

S S S S+ + = + + + ⋅ ≥� �+ + �

� (q.e.d)

2.6. Proposition 6: Letting that , , , , ,a b ca b c S S S are real numbers satisfying conditions:

i) a ≤ b ≤ c or a ≥ b ≥ c

ii) There exist α > 0 such that ( )22 1 0a c bS S S+ α + α + ≥

iii) 0 0

1 10 0

c b a b

c b a b

a b c b c b a b

S S S S

S S S S

� �− ≥ α − − ≥ α −� �� + ≥ ∨ + ≥� ��

α + α +� �+ ≥ + ≥� �α α� �

Then we have: ( ) ( ) ( )2 2 2 0a b cS b c S c a S a b− + − + − ≥ (1)

Proof

WLOG suppose a ≤ b ≤ c and

∃α > 0 such that ( )22 1 0a c bS S S+ α + α + ≥

Consider following possibilities:

� If b = c then (1) ⇔ ( ) ( )2 0 0b c b cS S a c S S+ − ≥ ⇔ + ≥ (is always true)

� If b ≠ c then (1) ⇔ ( ) ( )2 2

1 0a b cb a b aS S Sc b c b

− −+ + + ≥− −

⇔ ( ) ( ) 2 21 0a b cf t S S t S t= + + + ≥ for b atc b

−= ≥ α−

We have: ( ) ( ) ( ) ( ) ( )2b cf t f t S t S t� �− α = − α + + α + + α� �

( ) ( ) ( ) ( ) ( ) ( )1 1b c b c b b c b c bt S S t S S S t S S S S S� � � �= − α + + + α + α + ≥ − α + α + + α + α +� � � �

( ) ( ) ( )2 12 1 0, 0c b c btt S S S S t− α α +� �� = − α α + + α = + ≥ ∀ ≥ α >� � � �α α� �

It follows: ( ) ( ) ( )22 1 0a c bf t f S A S≥ α = + α + α + ≥

Equality occurs ⇔ ( )

( )

( )2 22 21 0 1 0a c b a c b

t b a c b

S A S S A S

� �= α − = α −� �⇔� �+ α + α + = + α + α + =� ��

and 0

1 0

c b

c b

a b c b

S S

S S

� − ≥ α −�� + ≥��

α +� + ≥� α�


Consequence: Letting that , , , , ,a b ca b c S S S are real numbers satisfying conditions:

i) a ≤ b ≤ c or a ≥ b ≥ c

ii) 4 0a c bS S S+ + ≥

iii) 0 0

2 0 2 0

c b a b

c b a b

a b c b c b a b

S S S S

S S S S

� �− ≥ − − ≥ −� �� + ≥ ∨ + ≥� ��

+ ≥ + ≥� ��

Then we have: ( ) ( ) ( )2 2 2 0a b cS b c S c a S a b− + − + − ≥

3. Theorem on the expression of S.O.S.

3.1. Theorem:

Supposing that two homogeneous permutation polynomials A, B have the same degree and

the same number of variables. If so, the difference of these two polynomials can be written in

the form of S.O.S., that is:

1 2 1 21 2 1 2... ...n n

n ncyc cyc

a a a a a aα βα α β β−� � = � − 2))(( jiij aaaP

In which 1 1... ...n n mα + + α = β + + β = and { }niiaa 1== . In this case, A and B are also said to

have S.O.S relation.

Proof

First of all, we prove the following lemma

Lemma: Supposing { }niiaa 1== and 1 2 ... n mα + α + + α = , then:

( )1 2 21 1 2 ... ( )nm

n ij i jcyc cyc

a a a a P a a aαα α− = −� � �

Proof: We will proof the lemma by inducing based on k, which is the number of elements

except 0 belonging to the set { } 1

ni i=

α ( mn, are constant).

If 1=k , the theorem is obviously true.

If 2k = , we have to prove the existence of the expression 21 1 2 ( )( )m t m t

ij i jcyc cyc

a a a P a a a−− = −� � �

This can be proved with notice that: 2))(,()( babaPbmabtmta tmtmm −=−−+ −

Indeed, it is first of all worth noticing that the equation ( )( ) 0n tf x tx m t mx= + − − = has one

double root which is 1, because ( ) ( )1 1 0f f ′= = . Therefore, ( )f x can be expressed in the

form ( ) ( )2( 1) , deg 2Q x x Q m− = −

Letting axb

= , then we have: ( ) ( )( ) 22( )m m m t m t ma ab f ta m t b ma b b Q a bb b

− −= + − − = − .


However, ( )2m ab Qb

− is a polynomial having 2 variables ,a b because Q is a ( )2n − -degree

polynomial. Supposing that our proposition is already true with k , the number of the elements

except for 0 in the set { } 1

ni i=

α , with 1k + , we can transform this into the case of k as follows:

( )1 2 1 2 1 21 3 11 2 1 1 2 2 1 2 1 2

1 2 1 3 11 2

... ...k kk k

a a a aa a a a a+ +

α +α α +α α αα α αα α

+ +

α + α − α + α= −

α + α

3 1 3 11 2 1 21 21 3 1 2 3 1

1 2 1 2

... ...k kk ka a a a a a

α+ +α α αα +α α +α

+ +α α

+α + α α + α

Note that with 2k = : ( ) ( )( )

1 2 1 2 1 221 1 2 2 1 2 1 2

12 1 21 2

a a a aH a a a

α +α α +α α αα + α − α + α= −

α + α, we have:

( )( ) 31 1 3 11 2 1 2 1 22 1 21 2 1 12 1 2 1 3 1 2 3 1

1 2 1 2

... ... ...k k kk k ka a a Q a a a a a a a a a

α+ + +α α α αα α α +α α +α

+ + +α α

= − + +α + α α + α

Therefore, 11 21 1 2 1... km

kcyc cyc

a a a a +αα α+−� �

3 31 2 1 22 1 11 212 1 2 1 1 3 1 2 3 1

1 2 1 2

( )( ) ... ...m k kk k

cyc cyc cyc cyc

Q a a a a a a a a a aα αα +α α +α+ ++ +

α α= − − + − +

α + α α + α� � � �

11 2

3 31 2 1 2

1 1 2 1

2 1 11 212 1 2 1 1 3 1 1 2 3 1

1 2 1 2

...

( )( ) ... ...

kmk

cyc cyc

m k m kk k

cyc cyc cyc cyc cyc

a a a a

Q a a a a a a a a a a a

+αα α+

α αα +α α +α+ ++ +

−

α α = − − + − + −� � � �α + α α + α � �

� �

� � � � �

On the other hand, two expressions 31 2 11 1 3 1...m k

kcyc cyc

a a a aαα +α ++−� � and 31 2 1

1 2 3 1...m kk

cyc cyc

a a a aαα +α ++−� �

can be written in the form of S.O.S, from which we have 11 21 1 2 1... km

kcyc cyc

a a a a +αα α+−� � can also

be written in the form of S.O.S.

According to the induction principle, the theorem has been proved.

Application: Using the lemma and following inequality, we get q.e.d.

1 2 1 2 1 2 1 21 2 1 2 1 2 1 1 1 2... ... ... ...n n n nm m

n n n ncyc cyc cyc cyc cyc cyc

a a a a a a a a a a a a a aα β α βα α β β α α β β − = − + −� � � � � �

� � � � � �

3.2. Comment: Generally speaking, most symmetrical or permutation inequalities with

polynomials (or fractions) of the same degree and having 3 variables can all be transformed

into 3-variable homogeneous permutation polynomials, i.e. can be written in the form of

S.O.S. Therefore, the method of S.O.S can be used to solve nearly all the 3-variable

homogeneous inequalities. The next question is: “How can we transform them into typical

form of S.O.S?”. Naturally, the algorism mentioned above is a useful suggestion to

transforming a polynomial into the typical form of S.O.S. However, what about other

algebraic form? To answer this question, let’s consider the following definition:


Two polynomials A and B are considered to have a “close” S.O.S relationship if expression

( )A B− can be written directly in the form of S.O.S without using intermediary expressions.

In this case, we can alter the expressions to clarify the S.O.S relationship via “close”

intermediary relationship.

Below are some common types of expressions:

i) ( , , ) ( , , ) ( , , ) ( , , )P a b c A a b c Q a b c B a b c− , in which A and B, P and Q are “close” to each other. To transform

the given expression into S.O.S form, we will use the intermediary quantity ( , , ) ( , , )Q a b c A a b c

[ ] [ ]( , , ) ( , , ) ( , , ) ( , , ) ( , , ) ( , , ) ( , , ) ( , , )

( , , ) ( , , ) ( , , ) ( , , ) ( , , ) ( , , )

P a b c A a b c Q a b c A a b c Q a b c A a b c Q a b c B a b c

A a b c P a b c Q a b c Q a b c A a b c B a b c

− + −

= − + −

ii) Or in the fractional form:

[ ]( , , ) ( , , ) ( , , )( , , ) ( , , ) ( , , ) ( , , )( , , ) ( , , ) ( , , ) ( , , ) ( , , )

B a b c Q a b c P a b cA a b c B a b c A a b c B a b cP a b c Q a b c P a b c P a b c Q a b c

−−− = +

iii) In root form:

( , , ) ( , , ) ( , , ) ( , , )( , , ) ( , , ) ( , , ) ( , , )

( , , ) ( , , ) ( , , ) ( , , )

A a b c P a b c B a b c Q a b cA a b c P a b c B a b c Q a b c

A a b c P a b c B a b c Q a b c

−− =+

iv) Specially cases:

2 2 22 2 2

3 3 3 2 2 2

( ) ( ) ( )2

( )3 ( ) ( ) ( )

2

a b b c c aa b c ab ac bc

a b ca b c abc a b b c c a

− + − + −+ + − − − =

+ +� �+ + − = − + − + −� �

3 3 32 2 2 2 2 2 ( ) ( ) ( )

3a b b c c a

a b b c c a ab bc ca− + − + −+ + − − − =

2 2 23 3 3 2 2 2

4 4 4 3 3 3

2 2 2 2 2 2 2 2 2

(2 )( ) (2 )( ) (2 )( )3

(3 2 )( ) (3 2 )( ) (3 2 )( )4

a b a b b c b c c a c aa b c a b b c c a

a b c a b b c c a

a ab b a b b bc c b c c ca a a c

+ − + + − + + −+ + − − − =

+ + − − −

+ + − + + + − + + + −=

3 3 3 3 3 3 3 3 3( ) ( ) ( )3

a b ca b b c c a ab bc ca b a c b a c

+ +� �+ + − − − = − + − + −� �

2 2 2 2 2 24 4 4 2 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) ( )

2a b a b b c b c c a c a

a b c a b b c c a− + + − + + − ++ + − − − =


4. Illustrative Problems

Problem 1. Prove that ( )( ) ( ) ( )2 2 2

1 1 1 94

ab bc caa b b c c a

� �+ + + + ≥� �+ + +� �, , , 0a b c∀ > (1)

(Iranian Mathematical Olympiad 1996)

Proof

WLOG supposing 0c b a≥ ≥ > .

Letting

2

2

2

b c x a x y z

c a y b x y z

a b z c x y z

+ = = − + +� �� + = ⇔ = − +� ��

+ = = + −� ��

� , ,x y z are the length of sides of a triangle.

Since 0c b a≥ ≥ > , 0x y z≥ ≥ > . The inequality (1) becomes:

( )2 2 22 2 2

91 1 12 2 244 4 4

xy yz zx x y zx y z

+ + − − − + + ≥� � �

( )2 2 22 2 2

1 1 12 2 2 9xy yz zx x y zx y z

⇔ + + − − − + + ≥� � �

⇔ 22

2 1( ) 0cyc

x yxy z

− − ≥� � �

� (4)

Letting 2 2 2

2 1 2 1 2 1, ,x y zS S S

yz zx xyx y z= − = − = −

The inequality (4) ⇔ 2 2 2( ) ( ) ( ) 0x y zS y z S z x S x y− + − + − ≥ (5)

As 0x y z≥ ≥ > and y z x+ > , so 0xS ≥ and 0yS ≥

We will prove 2 2 3 30y zy S z S y z xyz+ ≥ ⇔ + ≥ ⇔ ( ) ( )2 2y x y z yz xyz+ + − ≥ (6)

We have y z x+ > and 2 2y z yz yz+ − ≥ � (6) is true � 2 2 0y zy S z S+ ≥

Using Proposition 4 we have q.e.d.

Problem 2. Given , , 0a b c > be satisfying { }1min{ , , } max , ,4

a b c a b c≥ . Prove that

2

2 2 2 2

( )91 1 1 1( )4 16( ) ( ) ( ) ( )cyc

a bab bc ca

a b b c c a a b

−� �+ + + + ≥ +� �+ + + +� �

� (1)

Proof

WLOG supposing 1 04

c b a c≥ ≥ ≥ > . Letting

2

2

2

b c x a x y z

c a y b x y z

a b z c x y z

+ = = − + +� �� + = ⇔ = − +� ��

+ = = + −� ��

� , ,x y z are the length of sides of a triangle.

As 1 04

c b a c≥ ≥ ≥ > , so 0x y z≥ ≥ > and 4( ) 3 5 5x y z x y z y z x− + + ≥ + − � + ≥


The inequality (1) ⇔ ( )2

2 2 22 2 2 2

( )1 1 1 9 12 2 2

4 164 4 4 cyc

x yxy yz zx x y z

x y z z

− + + − − − + + ≥ +� � �

�

( )2

2 2 22 2 2 2

( )1 1 1 12 2 2 9

4 cyc

x yxy yz zx x y z

x y z z

− ⇔ + + − − − + + ≥ +� � �

� 22

2 5( ) 0

4cyc

x yxy z

⇔ − − ≥� � �

�

Letting 2 2 2

2 5 2 5 2 5, ,

4 4 4x y zS S Syz zx xyx y z

= − = − = −

We have to prove 2 2 2( ) ( ) ( ) 0x y zS y z S z x S x y− + − + − ≥

As 0x y z≥ ≥ > and 3 5 5y z x+ ≥ , so 0xS > and 8 5y x≥ � 0yS ≥

We will prove ( )2 2

2 2 3 32 2 50 4 5

2y zy z

y S z S y z xyzxz xy

+ ≥ ⇔ + ≥ ⇔ + ≥

We have 3 5 5y z x+ ≥ , it is suffices to prove that 3 34( ) (3 5 )y z y z yz+ ≥ +

2 2( )(4 4 ) 0y z y yz z⇔ − + − ≥ (is always true). Thus 2 2 0y zy S z S+ ≥

We have ( ) 0y

x z x yz

− ≥ − ≥ . Hence, 2 2 2 2 2( ) ( ) ( ) ( ) ( )x y z y zS y z S z x S x y S z x S x y− + − + − ≥ − + −

2 2 222 2

2 2

( ) ( ). .( ) ( ) 0y z

y z

x y y S z SyS x y S x y

z z

− +≥ − + − = ≥ � q.e.d.

Equality occurs ⇔ x y z= = or 58

y z x= = .

Problem 3. Given 3 positive real numbers a, b, c are satisfying 1abc ≥ . Prove that

S = 5 2 5 2 5 2

5 2 2 5 2 2 5 2 20a a b b c c

a b c b c a c a b− − −+ + ≥

+ + + + + +

(IMO 45 −−−− 2005)

Proof

Since 1abc ≥ , ( ) ( ) ( )5 2 5 2 5 2

5 2 2 5 2 2 5 2 2a a abc b b abc c c abcS

a b c abc b c a abc c a b abc− − −≥ + +

+ + + + + +

( )( )

( )4 2 2 24 2

24 2 2 4 2 2

2

2cyc cyc

a a b ca a bca b c bc a b c

− +−= ≥+ + + +

� �

Letting 2 2 2, ,x a y b z c= = = ( )

( )( )

( )( )

( )

2 22

2 2 222 2

2 22

22 2

x x y z z z x yy y z xS

y z xx y z z x y

− + − +− +� ≥ + +

+ ++ + + +

( )( ) ( )

( ) ( )( ) ( )

2 2 22

2 2 2 222 2 2

20

22 2 2cyc cyc

x y xy z z x yyxx y x yy z xx y z x y z y z x

+ − + + + = − − = − ≥� �� + ++ + + + + + � � ��

Equality occurs ⇔ x y z a b c= = ⇔ = =


Problem 4. Find the maximum coefficient k (k > 0) such that

2 2 232

a b c ab bc cak kb c c a a b a b c

+ ++ + + ≥ ++ + + + +

, , , 0a b c∀ > (1)

Solution

By a general correspondence: (1) ⇔ ( ) ( )2 2 21 12cyc

a ab bc cakb c a b c

+ +− ≥ −+ + +� (2)

We have: ( ) ( ) ( ) ( ) ( )1 1 1 1 12 2 2cyc cyc cyc

a b a ca a bb c b c b c c a

− + −− = = − −+ + + +� � �

( )( ) ( )

212 cyc

a bb c c a

−=+ +�

(2) ⇔ ( )

( ) ( )( )2 2 2 2

2 2 22

cyc

a b c ab bc caa b kb c c a a b c

+ + − − −− ≥+ + + +�

( )( ) ( )

( )2 2

2 2 2cyc cyc

a b a bkb c c a a b c

− −⇔ ≥+ + + +� �

⇔ ( )

( ) ( )( ) ( )

( ) ( )

2 2 2 2 22

2 2 20 0

cyc cyc

a b a b a b ck a b kb c c a b c c aa b c

� � � �− − + +− ≥ ⇔ − − ≥� � � �+ + + ++ + � �� (3)

The necessary condition:

Taking c = b ≠ a into (3) � ( )( )

2 22 2 02a ba b kb a b

� �+− − ≥� �+� � � ( ) ( )

2 22 ,2a bk u a bb a b

+≤ =+

We have: ( )( )

( )2

222,

2 22 1

ab tu a b

tab

++= =++

� ( ) ( )2 22 2 2 2 2 2 0t u t f t t ut u+ = + ⇔ = − + − =

u ∈ the domain of the function ⇔ ƒ(t) = 0 has root ⇔ 2 2 2 0t u u′∆ = + − ≥ � 3 12

u −≥

The sufficient condition:

We will prove ( )3 1 Min ,2

k u a b−= = is the best coefficient of inequality (1)

Indeed, with 3 12

k −= we have: (1) ⇔ ( )( ) ( )

2 2 22 0cyc

a b ca b kb c c a

� �+ +− − ≥� �+ +� �� (4)

Letting ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 2

; ;a b ca b c a b c a b cS k S k S kc a b a a b c b b c a c

+ + + + + += − = − = −+ + + + + +

WLOG supposing a b c≥ ≥ � a b cS S S≤ ≤

Consider ( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )2 2 22 2 2 2 2 2 22 2a b

a b c a b ca b c a b cS S k kc a b a a b c b a b b c c a

+ + + ++ + + ++ = + − = −+ + + + + + +

Letting 2

a bv += �

( ) ( )

( ) ( )( ) ( )( ) ( ) ( )

22

2 2 2 2

2

2 222 22 2 2

2

2

a b

a b c v cv c v c v cS S k k k

v b c c a v v cb c c av

� �+ + +� � + +� � ++ ≥ − ≥ − = −+ + +� �+ + +

� ��

� ( ) ( )2 222 2 , 0

2a bv cS S k u v c kv v c

+ � �+ ≥ − = − ≥� � � �+ � (since ( ) 3 1,

2u v c k−≥ = )


As a b cS S S≤ ≤ and 0a bS S+ ≥ , so 0; 0; 0b b c c aS S S S S≥ + ≥ + ≥

From Proposition 2 it follows: with 3 12

k −= then (4) is true ∀a, b, c > 0.

Conclusion: 3 12

k −= is the maximum coefficient to make the inequality (1) is true ∀a, b, c > 0.

Problem 5. (Pham Kim Hung) Given , , 0a b c ≥ . Determine the best constant for the

following inequality: ( ) ( ) 2 2 21 1 1 9

ab bc caa b c k k

a b c a b c+ ++ + + + + ≥ ++ +

(1)

Solution

We use the following change:

( ) ( ) ( ) ( ) ( )

( ) ( ) ( )( )

2 2 2

2 2 2

2 2 2 2 2 2

1 1 1 9

12

b c c a a ba b c

a b c bc ca ab

ab bc ca b c c a a b

a b c a b c

− − −+ + + + − = + +

+ + − + − + −− =+ + + +

Then the inequality (1) becomes: ( )( ) ( )

22

2 2 2

10 0

2a

cyc cyc

kb c S b c

bc a b c

− − ≥ ⇔ − ≥� �+ + �� (2)

In which: ( ) ( ) ( )2 2 2 2 2 2 2 2 22 , 2 , 2a b cS a a b c kabc S b a b c kabc S c a b c kabc= + + − = + + − = + + −

Now, it is given that b c= , then (2) becomes:

( ) ( ) ( )22 20 2 2 0 2 2 4 2 0bS a b kab a b k ab≥ ⇔ + − ≥ ⇔ − + − ≥

Taking 2a b= then 4 2k ≤ . We shall prove the inequality (2) with 4 2k = .

Supposing a b ca b c S S S≥ ≥ � ≥ ≥ . On the other hand:

( ) ( ) ( ) ( )22 2 2 22 8 2 4 . 2 8 2 4 2 0b cS S b c a b c abc bc a bc abc bc a bc+ = + + + − ≥ + − = − ≥

Using the proposition 2, we got the q.e.d.

Conclusion: 4 2k = is the best constant of the inequality.

Problem 6. (Pham Kim Hung) Given , , 0a b c ≥ . Determine the best constant for the

following inequality: 2 2 2 2 2 2 2 2 2

35

bc ca abb c ka c a kb b a kc

+ + ≤+ + + + + +

(1)

Solution

Given that 3, 2, 2a b c= = = , it follows: 3k ≤ . We shall prove the inequality when 3k = .

Indeed, WLOG, supposing a b c≥ ≥ and 2 2 2 1a b c+ + = . We use the following change:

( ) ( ) ( )22 2 2 2 2 2 22 3 10 5 3b c a bc b c a b a c+ + − = − + − + −


Thus the inequality (1) ⇔ ( ) ( ) ( )

2 2 2 2 22

2

5 30 0

1 2 acyc cyc

b c a b a cS b c

a− + − + − ≥ ⇔ − ≥

+� �

In which: ( ) ( ) ( ) ( )22 2 25 1 2 1 2 6 1 2aS b c b c a= + + − + + ;

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 22 2 2 2 2 25 1 2 1 2 6 1 2 ; 5 1 2 1 2 6 1 2b cS c a c a b S a b a b c= + + − + + = + + − + +

Let 2 2

2a b

x+= . Obviously, 0cS ≥ and ( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 21 2 1 2 2 1 2a c b b c a x c x+ + + + + ≤ + +

We shall prove 0a bS S+ ≥ . Indeed:

( ) ( ) ( ) ( ) ( )2 22 2 210 1 2 1 2 6 1 2 0 2 3 0a bS S c x x c x x c+ ≥ + + − + + ≥ ⇔ − ≥ (true)

Next we shall prove 0bS ≥ . Indeed: ( ) ( ) ( ) ( )22 2 25 1 2 1 2 6 1 2 0bS c a c a b= + + − + + ≥

( )( )

( )( )

22 2 2 2 2 2 2 2

2 2 2 2 2 22 2 2 2 2 2

3 6 2 12 5 9

3 35 3 5 3

a b c a c a b a ac b c

b c a b c ac a b c a b

+ + + − + − −⇔ ≥ ⇔ ≥+ + + ++ + + +

This is obviously true as: ( ) ( )2 2 2 2 2 2 2 2 2 2 22 12 5 9 , 3 5 3a b a ac b c b c a c a b− ≥ + − − + + ≤ + +

Using the proposition 2, we got the q.e.d.

Conclusion: k = 3 is the best constant of the inequality.


1

a b c

ab bc ca

>��

+ + =��. Prove that

( ) ( ) ( )2 2 2 2 2 2

2 2 21 1 1 5

2a b b c c a

a b b c c a+ + ++ + ≥

+ + + (1)

Proof

The inequality (1) ⇔ ( )

( )( )

2 2 2

252cyc

ab bc ca a b ab bc caa b

+ + + ≥ + ++

�

⇔ ( ) ( )

( )( )

2

222 5

cyc

ab ab bc ca bc ca ab bc caa b

+ + + + ≥ + ++

�

⇔ ( )( ) ( ) ( )

( ) ( )2 2 22 2 2

4 2 5ab bc caab bc ca a b c ab bc caa b b c c a

� �+ + + + + + + ≥ + +� �+ + +� �

⇔ ( )( ) ( ) ( )

( )2 2 22 2 2

4 4 4 3 2 0ab bc caab bc ca a b c ab bc caa b b c c a

� �+ + + + − + + + − − − ≥� �+ + +� �

⇔ ( ) ( )( )

( )( )

( )( )

( ) ( ) ( )2 2 2

2 2 2

2 2 20a b b c c aab bc ca a b b c c a

a b b c c a

� �− − −− + + + + + − + − + − ≥� �+ + +� �

⇔ ( )

( )( )

( )( )

( )2 2 2

2 2 21 1 1 0ab bc ca ab bc ca ab bc caa b b c c a

a b b c c a+ + + + + +� � � � � �− − + − − + − − ≥

� � � � � �+ + +� � � � � �

Letting ( ) ( ) ( )2 2 2

1 ; 1 ; 1a b cab bc ca ab bc ca ab bc caS S S

b c c a a b+ + + + + += − = − = −

+ + +


WLOG supposing a ≤ b ≤ c � a b cS S S≥ ≥

We have: ( )

( ) ( )( )

2

2 21 0b

a a c c bab bc caSc a c a

+ + −+ += − = >+ +

� 0a bS S≥ >

On the other hand, ( ) ( )

2 2 2 23 2 2 2

1 1ab bc ca ab bc cab A c A b ca b c a+ + + +� � � �+ = − + −

� � � �+ +� � � � =

( ) ( )( )

( ) ( )( ) ( ) ( )

( ) ( )2 2 2 2 2 22 2 2

2 2 2 2a a b b c a a c c b b c c bb c a c b

a c a ba b c a a b c a

+ + − + + − � �= ⋅ + ⋅ = + + − −� � + ++ + + +� �

( ) ( )( )

( ) ( )2 2

22 2

0b c ab bc caa c ba c a ba b c a

2� � + += + + − ⋅ >� � + ++ +� �

Using Proposition 4 we have (q.e.d). Equality occurs ⇔ 13

a b c= = =

Problem 8. Letting that , ,a b c be length of sides of a triangle. Prove that

{ } ( ) ( )1 1 13Min ,a b c b c a a b cb c a a b c a b c

+ + + + ≥ + + + + (1)

Proof

WLOG supposing { }Min ,a b c b c a a b cb c a a b c b c a

+ + + + = + + , then we have:

(1) ⇔ ( ) ( ) ( )1 1 13 a b c a b cb c a a b c

+ + ≥ + + + + ( ) ( ) ( )2 2 23 a c b a c b a b c ab bc ca⇔ + + ≥ + + + +

( ) ( )2 2 2 2 2 22 3a c b a c b a b b c c a abc⇔ + + ≥ + + +

( ) ( )3 3 3 2 2 22 a b c a c b a c b� �⇔ + + − + +� � ( ) ( ) ( )3 3 3 2 2 2 3 3 3 3a b c a b b c c a a b c abc≤ + + − + + + + + − (2)

We have:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

23 3 2 2 2

23 3 2 2 2

23 3 2 2 2

2 3 2 2

2 3 2 2

2 3 2 2

b a b a b a b ba a b a b a

c b c b c b c cb b c b c b

a c a c a c a ac c a c a c

� + − = − − − = − +��

+ � + − = − − − = − +�

+ − = − − − = − +��

� ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 23 3 3 2 2 23 2 2 2a b c a c b a c b a b b a b c c b c a a c� �+ + − + + = − + + − + + − +� �

Similarly, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 23 3 3 2 2 23 2 2 2a b c a b b c c a a b a b b c b c c a c a� �+ + − + + = − + + − + + − +� �

On the other hand, ( ) ( ) ( ) ( )2 2 23 3 3 132

a b c abc a b c a b b c c a� �+ + − = + + − + − + −� �

It follows: (2) ⇔ ( ) ( ) ( ) ( ) ( ) ( )2 2 22 1 12 23 3 2cyc cyc cyc

b a a b a b a b a b c a b+ − ≤ + − + + + −� � �

⇔ ( ) ( ) ( ) ( ) ( ) ( )2 21 1 2 12 2 0 03 2 3 2cyc cyc

a b a b a b c b a c a b a b� �− + + + + − + ≥ ⇔ + − − ≥� ��

⇔ ( ) ( ) ( ) ( ) ( ) ( )2 2 2 0c a b a b a b c b c b c a c a+ − − + + − − + + − − ≥ (3)


Letting ; ;a b cS a b c S b c a S c a b= + − = + − = + −

� 2 0 ; 2 0 ; 2 0a b b c c aS S b S S c S S a+ = > + = > + = > . Beside, we have:

( ) ( ) ( ) ( ) ( ) ( )a b b c c aS S S S S S a b c b c a b c a c a b c a b a b c+ + = + − + − + + − + − + + − + −

= ( ) ( ) ( ) ( ) ( )2 2 22 2 2 2 2 22a b c b c a c a b ab bc ca a b c− − + − − + − − = + + − + +

= ( ) ( ) ( )24 2 2ab a b c ab a b c ab a b c− + − = + + − − − +

= ( ) ( ) ( ) ( )a b c a b c b c a c a b+ + + − + − + −

Since , ,a b c are length of sides of a triangle, it follows 0a b b c c aS S S S S S+ + > .

Using Proposition 5 we have (3) is true � (2) is true � (1) is true.

Problem 9. (Vasile Cirtoaje) Letting that a, b, c are the length of 3 sides of a triangle.

Prove that ( ) ( )3 2 3a b c a c bb c a c b a

+ + ≥ + + + (1)

Proof

The inequality (1) ⇔ ( ) ( )2 2 2 2 2 23 2 3a c b a c b b c c a a b abc+ + ≥ + + +

⇔ ( ) ( ) ( ) ( ) ( )3 3 3 2 2 2 3 3 3 2 2 2 3 3 33 2 3a b c a c b a c b a b c a b b c c a a b c abc� � � �+ + − + + ≤ + + − + + + + + −� � � �

⇔ ( ) ( ) ( ) ( ) ( ) ( )2 2 22 12 23 2cyc cyc cyc

b a a b a b a b a b c a b+ − ≤ + − + + + −� � �

⇔ ( ) ( ) ( ) ( ) ( ) ( )2 2 25 5 3 5 5 3 5 5 3 0a b a b c b c b c a c a c a b− − + + − − + + − − + ≥

Letting 5 5 3 ; 5 5 3 ; 5 5 3a b cS b c a S c a b S a b c= − + = − + = − +

WLOG suppose { } { }in , , , ,M a b c b Max a b c≤ ≤ . Consider 2 following possibilities:

� c ≥ b ≥ a: We have 5 5 3 0; 8 2 0; 8 2 0b a b b cS c a b S S b a S S c b= − + > + = − ≥ + = − ≥


� a ≥ b ≥ c: We have ( )4 18 12 12 12 0a c bS S S c b a c b a+ + = + − > + − >

++++ If a b c b− ≥ − then c > a − b ≥ b − c ≥ 0 � 2c > b

� 8 2 4 2 2 0b cS S c b b b b+ = − > − = > � ( )2 2 5 0c b bS S S b c a b+ > + = − + >

++++ If c b a b− ≥ − then b − c ≥ a − b ≥ 0 � 2b ≥ a + c > a

� 8 2 4 2 2 0a bS S b a a a a+ = − > − = > � ( )2 2 5 3 3 3 0a b bS S S a c a b c a b+ > + = − + > − + >

Using Proposition 6 with α = 1 we have q.e.d.

Problem 10. Given , , 0x y z > . Prove that

( )2 2 2 2 2 2 2 2 22x y z xy yz zx x y z y z x z x y+ + + + + ≥ + + + + + (1)

Proof

(1) ⇔ ( )2 2 2 2 2 2 2 2 22( ) 2( ) 2 2 4( )x y z xy yz zx x y z y z x z x y xy yz zx+ + − + + ≥ + + + + + − + +


( )2 2 2

2 22 2

2 2 2 2

( ) 2 2( ) ( )

( ) ( )( ) 2 ( ) 2

2( ) 2( )

cyc cyc

cyc cyc cyc cyc

x y x y z x y z

x y z z x yx y x y

y z y z x y x y

⇔ − ≥ + − +

− −⇔ − ≥ ⇔ − ≥+ + + + + +

� �

� � � �

Using Cauchy-Schwarz inequality, we have: 2 2

2 2

( ) ( )2

2( )cyc cyc

z x y z x yx yx y x y

− −≤++ + +

� �

So it is suffices to prove that: 2

2 ( )( )

cyc cyc

z x yx y

x y−− ≥+� � 21 ( ) 0

cyc

z x yx y

⇔ − − ≥� �+ �� (2)

Letting 1 , 1 , 1x y z

yx zS S S

y z z x x y= − = − = −

+ + +, then (2) ⇔ 2 2 2( ) ( ) ( ) 0x y zS y z S z x S x y− + − + − ≥

WLOG supposing 0x y z≥ ≥ > , then , 0y zS S >

We have 2 2

2 2 2 2 2 2 2 0y xx y xy

x S y S x y x y xyx z y z

+ = + − − ≥ + − ≥+ +

Using Proposition 4 we have q.e.d. Equality occurs ⇔ x y z= = .

Problem 11. (Komal) Letting that , , 0a b c > are satisfying 1abc = . Prove that

2 2 2 2 2 2

1 1 1 3 1 1 1 2a b c a b c a b c a b c

+ + − ≥ + +� �+ + + + � (1)

Proof

The inequality (1) 2 2 2 2 2 2

2 2 2

2( )3 a b b c c aabcab bc ca

a b c a b c+ +⇔ + + − ≥

+ + + +

2 2 2 2 2 2

2 2 2

2( )9 6a b b c c aabc abcab bc ca

a b c a b ca b c+ +⇔ + + − ≥ −

+ + + ++ +

2 2 22 2 2

2 2 2

( ) ( ) ( 2 )( ) ( 2 ) 0

( )( )cyc cyc cyc

c a b c a b c bc ca abc a b a b ab bc ca

a b c a b c a b c

− − + + −⇔ ≥ ⇔ − + + − − ≥+ + + + + +� � � (2)

Letting 2 2 2 2( 2 ) ; ( 2 ) ;a bS a b c bc ca ab S b c a ca ab bc= + + − − = + + − −

2 2( 2 )cS c a b ab bc ca= + + − − . Then (2) ⇔ 2 2 2( ) ( ) ( ) 0a b cS b c S c a S a b− + − + − ≥

WLOG supposing a b c≥ ≥ , it easily see that , 0b cS S ≥ .

We have 2 2 2 2 2 2 2( ) ( ) 2 ( ) ( ) 0b aa S b S ab a b a b c a b ab c a b� �+ = − + + + − + + >� �


Problem 12. (Vo Quoc Ba Can)

Given , ,x y z satisfy 1xy yz zx+ + = . Prove that 2 2 2

2 2 22( ) 3 2x y z

x y zy z x

+ + − + + ≥ − (1)


Proof

The inequality (1) ⇔ 22 2

2 2 23 2( 1)yx z x y z x y z x y z

y z x

+ + − − − + + + − ≥ + + −� � �

2 22( ) ( )1

( )2 3cyc cyc cyc

x y x yx y

y x y z

− −⇔ + ≥ −+ + +

� � � ( )2 1 1( ) 1 0

2 3cyc

x yy x y z

� �⇔ − + − ≥� �+ + +� �

� (2)

Letting ( ) ( )1 1 1 1

1 ; 1;2 3 2 3

x yS Sz xx y z x y z

= + − = + −+ + + + + + ( )

1 11

2 3zS

y x y z= + −

+ + +

Now, the inequality (2) ⇔ 2 2 2( ) ( ) ( ) 0x y zS y z S z x S x y− + − + − ≥

We have ( )1 1 1 3

32 3

x y zS S Sx y z x y z

+ + = + + − ++ + +

( )3

32 3

xy yz zxxyz x y z

+ += − ++ + +

( )1 3

32 3xyz x y z

= − ++ + +

( ) ( )32

3 3 3 33 3 3 3 0

2 3 2 3( ) x y z x y zxy yz zx≥ − + = − + >

+ + + + + ++ +

Letting ( )1

2 3t

x y z=

+ + +, we have:

( ) ( ) ( ) ( )1 1 1 1 1 11 1 1 1 1 1x y y z z xS S S S S S t t t t t tx y y z z x

+ + = + − + − + + − + − + + − + −� � � � � �

2 1 1 11 1 1 1 1 13 2 3 2 3

3 21 1 1 1 1 12 3

t tx y z x y zxy yz zx

x y z xyzx y zxy yz zx xyz

= + + + − + + + − + + +� � � � � �

+ + + − > + + − + + + =� � �

We will prove 3 2

0 3 2 0x y z xyz

x y z xyzxyz

+ + + − ≥ ⇔ + + + − ≥ (3)

If 2x y z+ + ≥ then (3) is obviously true.

If 2x y z+ + ≤ , letting p x y z= + + � 2 3p≥ ≥ .

Now using Schur inequality, we have:34

9p p

xyz−≥ . Hence:

3 34 7 6 (2 )( 1)( 3)3 2 2 0

3 3 3p p p p p p p

p xyz p− − + − − − ++ − ≥ − + = = ≥ � (3) is true

Thus we have 0

0

x y z

x y y z z x

S S S

S S S S S S

+ + >��

+ + >��

, so 2 2 2( ) ( ) ( ) 0x y zS y z S z x S x y− + − + − ≥ � (q.e.d)


x y z= = =


Problem 13. (Nguyen Van Thach)

Given , , 0a b c > . Prove that 2 2 2 4 4 4

2 2 23

a b c a b cb c a a b c

+ ++ + ≥ ⋅+ +

(1)

Proof

The inequality (1) ⇔ 4 4 44 2

2 2 2 2

9( )2

cyc cyc

a b ca a bcb a b c

+ ++ ≥+ +� �

4 4 44 22 2 2 2

2 2 2 2

2 22

2 2 2 22 2 2

9( )2 2 2 3 2 2

3 ( ) ( )

( ) 1 2 ( ) ( ) ( ) 0

cyc cyc cyc cyc cyc

cycc

cyc cyc cyc cyc

a b ca a bb a bc ab a a ab

cb a b c

a b a ba b

a b c a a b S a bb c a b c

+ + ⇔ + − + + − ≥ − + −� �� + + � � �

− + ⇔ − + + − ≥ + − ⇔ − ≥� � � + +

� � � � �

��

in which 2 2 2 2

2 2 2 2 2 2 2 2

2 2 3( ) 2 2 3( );a b

b b a b c c c b c aS S

c b a cc a b c a a b c+ += + + − = + + −

+ + + +

22

2 2 2 2

3( )2 2c

a ba a cS

b ab a b c+= + + −

+ +

Obviously, we just need to consider inequality in case 0a b c≥ ≥ >

* Case 1. 2( ) 2b c a b b c a c b a c− ≥ − ⇔ − ≥ − ⇔ ≥ + . We have:

2 2 2 2

2 2 2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2 2 2 2 2

4 2 2 3( ) 3( )

4 3( ) 3( )4

3( ) 3( ) 7 4 7 6 610 0

a ca b a b c a b b c

S Sb c ab c a b c

a b a b a b b cb ab c a b c

a b b c a b c ab bca b c a b c

+ + ++ = + + + + −+ +

+ + +≥ + + + −+ +

+ + + + + − −≥ − = ≥+ + + +

2 22 2

2 2 2 2 2

2 22 2

2 2 2 2 2 2

3( ) 3( )2 2 2 2

3 (2 ) ( 2 )3( ) 3( )9 0

2( )

a ba c b cb b a c c b

S Sc b a cc a a b c

a b b ca c b c

a b c a b c

+ + ++ = + + + + + −+ +

� �− + −+ + + � �≥ − = ≥+ + + +

2 2 2 2

2 2 2 2 2

2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 2 2 2

4 8 10 2 3( ) 12( )4

4 8 5 2 3( ) 12( )1 5

3( ) 12( ) 10 19 7 24 622 0

a bc c b a b b c a c

S Sa c ba c a b c

c c a b a b c a ca a c bc a b c

b c a c a b c ac bca b c a b c

+ + ++ = + + + + −+ +

+ + +≥ − + + + + + −+ +

+ + + + + − −≥ − = ≥+ + + +


2 2 22 2 2

2 2 2 2 2 2

2 2 2

2 2 2

2 2 2

2 2 2

3( ) 12( ) 3( )4 10 10 44

3( ) 12( ) 3( )26

11 20 11 240

a b cb c a c a bc c b a b a

S S Sa c ba c b a b c

b c a c a b

a b c

a b c aba b c

+ + + + ++ + = + + + + + −+ +

+ + + + +≥ −+ +

+ + −= ≥+ +

If 0bS ≥ then we have 2 2( ) ( )( ) 0c a ccyc

S a b S S a b− ≥ + − ≥�

If 0, 0b cS S≤ ≥ then 2 2( ) ( 4 )( ) 0c a bcyc

S a b S S b c− ≥ + − ≥�

If , 0b cS S ≤ then 2 2( ) ( 4 )( ) 0c a b ccyc

S a b S S S b c− ≥ + + − ≥�

* Case 2. .a b b c− ≥ − We will prove 0cS ≥

Indeed, consider the function 2 2

2 2 2 2

2 2 3( )( ) c

a a c a bf c S

b ab a b c+= = + + −

+ +

Clearly, this function is increasing with c , so

+ If 2b a≤ , we have: 2 2 2 2 2

2 2 2 2 2 2 2

2 3( ) 3( ) 5 5 6( ) (0) 8 0

a a a b a b a b abf c f

bb a b a b a b+ + + −≥ = + − ≥ − = ≥+ + +

+ If ,2 ab ≥ we have: 2 2

2 2 2

2 4 3( )( ) (2 ) 2 0

2 5 4a a b a b

f c f b ab ab a b ab

+≥ − = + + − − ≥+ −

Thus , 0a cS S ≥ . So if 0bS ≥ then we right have q.e.d.

Consider case 0bS ≤ , we have:

2 2 2 2

2 2 2 2 2

2 2 2 2

2 2 2 2 2 2

2 4 6 2 6( ) 3( )2

8 8 2 6( ) 3( ) 6( ) 3( )2 1 12 0

a bc c b b a a c b c

S Sa c ba c a b c

c b a a c b c a c b ca c b a b c a b c

+ + ++ = + + + + −+ +

+ + + + + +≥ − + − + − ≥ − ≥+ + + +

( )

2 22 2

2 2 2 2 2

2 2 2 2

2 2 2 2 2 2

6( ) 3( )2 6 4 22

2 2 6 6( ) 3( ) 6( ) 3( )4 41 1 13.6 0

c ba c a bc c b a a

S Sa c ba b a b c

c a c a b a c a bb aa c b a b c a b c

+ + ++ = + + + + −+ +

+ + + + + + +≥ − + + − − ≥ − ≥+ + + +

� 2 2 2( ) ( 2 )( ) ( 2 )( ) 0c a b c bcyc

S a b S S b c S S a b− ≥ + − + + − ≥�

The inequality is proven. Equality occurs ⇔ .a b c= =


Problem 14. (Nguyen Van Thach)

Prove that 2 2 2 2 2 2 2 2 2

2a b c a ab b b bc c c ca ab c a a b b c c a

− + − + − ++ + ≥ + +� �+ + + � , , , 0a b c∀ > (1)

Proof

The inequality (1) ⇔ 2 2 2

( ) 22cyc cyc

a a ab b a b ca b c

b a b

− + + +− + + ≥ −� �� + ��

2 22( ) ( )3 1 3

( ) 02 2( )cyc cyc cyc

a b a ba b

b a b b a b − −⇔ ≥ ⇔ − − ≥� �+ + �

� � �

Letting 1 3 1 3 1 3

; ;2( ) 2( ) 2( )a b cS S S

c b c a c a b a b= − = − = −

+ + +

WLOG we just need to consider the case cba ≥≥ . Then we have .0, ≥ca SS

So if 0≥bS then we right have q.e.d.

Consider ,0≤bS we have: 2 3 1 3 3 3 3 3

2 02( ) 4 2( )a bS S

a a c c b c a c a c c b c+ = − + − ≥ − + − ≥

+ + + + +

2 1 3 3

22( )c bS S

a b a c a b+ = + − −

+ +

+ Case 1. 2( ) 2b c a b b c a c b a c− ≥ − ⇔ − ≥ − ⇔ ≥ +

We have ( )4 1 6 3 4 1 6 32 4 1432( ) 3 3 3a bS S

a ca c a c b c a c a c a c+ = + − − ≥ + + + − −

+ + + +

22 12 9 6 33

03( ) 3 3a c a c a c a c

+� �⋅ �≥ + − − ≥+ + + +

. Then 2 2( ) ( 4 )( ) 0c a bcyc

S a b S S b c− ≥ + − ≥�

* Case 2. 2b c a b a c b− ≤ − ⇔ + ≥ . Letting ( ) 2c bf c S S= + . Clearly, ƒ(c) is increasing, so

+ If 2b a≤ then 2 22 3 1 3 3 2 3 2

( ) (0) 02( ) 2( ) 2 ( )

a b a ab bf c f

a a b a b ab a b ab a b− − −≥ = − + − = − = ≥

+ + +

+ If 2b a≥ then 2 3 1 3 2 1 3

( ) (2 ) 02 2( ) 2

f c f b aa b b a b a a a a

≥ − = − + − ≥ − − =+ +

Thus ƒ(c) ≥ 0, so we have: 2 2 2( ) ( 2 )( ) ( 2 )( ) 0c a b c bcyc

S a b S S b c S S a b− ≥ + − + + − ≥�

The inequality is proven. Equality occurs ⇔ .cba ==

Comment: We can see the use of S.O.S principle in solving three variables inequality, but

what about multi-variable ones? In fact, S.O.S proves to have little help, but we still can

apply it in many cases, and it is usually accompanied by the induction principle and mixing

variables technique. Let’s consider some examples to clarify this problem.


Problem 15 (Vasile Cirtoaje)

Giving n positive real numbers 1 2, ,..., na a a satisfying 1 2 ... na a a n+ + + = . Prove that:

2 2 2 2 21 2

1 2

1 1 1... 4( 1)( ... ) ( 2)nn

n n a a a n na a a

+ + + ≥ − + + + + −� � �

(1)

Proof

We will prove the inequality by induction. And here we just consider the most complex part,

that is proving the inequality with 1n + numbers when we already knew that it is true with n

or less than n numbers.

Firstly, notice that the inequality is equivalent to:

2 2 2 2 2 21 1

1 2

1 1 1( ... ) ( 2) ( ... ) 4( 1) ( ... )n n

n

n n n a a n n a aa a a

� �+ + + − − + + ≥ − + +� �

� �

So if the inequality is true for 1 ... na a n+ + = then it is also true for 1 ... na a n+ + ≤ .

Letting ( )1 2, ,... nf a a a LHS RHS= − .

WLOG supposing { }1 1 2max , ,..., na a a a= . We have:

2 21 1

222 2 2 2

22 2

... ...( ,..., ) , ,...,

1 1

( ... )1 1 4( 1) ...... 1

n nn

nn

n n

a a a af a a f a

n n

a ann n a aa a a a n

+ + + + − � �− − �

� �+ + = + ⋅ ⋅ ⋅ + − − − + + −� �� + + −� � �

2 22 22 2

2

( ) ( )4 ( ) 4 ( )

( ... ) 1i j i j

i j i ji j i jn

a a a an na a a a

a a a aa a n

− −= − − ≥ − −

+ + −� � � � (2)

However, we have already had the inequality is true with n variables:

2 2 2 22

2

1 1( 1) 4( 2)( ... ) ( 1)( 3)nn

n n a a n na a

− + ⋅ ⋅ ⋅ + ≥ − + + + − −� � �

2 222 2

( ) ( )4( 2) ( 1)( 1) ( ) 4 ( )

1 2i j i j

i j i ji j i j

a a a an nn a a a a

a a a an n

− −− −⇔ − ≥ − ⇔ ≥ −− −� � � �

Clearly, 2 2( 1)1 2

n nn n

−>− −

, so 22

2( )

4 ( ) 01

i ji j

i j

a ana a

a an

−− − ≥

− � �

Therefore, we only have to prove that ( )( , , ..., ) 0, 1f a d d a n d n≥ + − = . This is not so

difficult and you can easily find out the answers. The application technique is also the

performance of S.O.S.


Problem 16 (Vasile Cirtoaje)

Given n positive real numbers 1 2, ,..., na a a . Prove that

( )1 1 1 1

1 2

1 1 1... 1 ... ... ( ... ) ...n nn n n n

n

a a n n a a a a a aa a a

+ + + − ≥ + + + + +� � �

Proof

The proof below uses Suranyi inequality:

( )1 11 1 2 1 1( 1)( ... ) ... ... ...n n n n n n

n n n nn a a a a a a a na a− −− + + − + + ≥ + + −�

We will again prove the inequality by induction. And here we just consider the most complex

part, that is proving the inequality with 1n + numbers when we already knew that it is true

with n or less than n numbers.

The inequality ⇔ ( ) 21 1 2 1 1

1 2

1 1 1... ... ... ( ... )n nn n n n

n

a a na a a a a a a na a a

� � + + − ≥ + + + + ⋅ ⋅ ⋅ + −� ��

2

1 1 2

( )( ... ... ) .. i jn n

n n i ji j

a aa a na a a a a

a a

� �−� �⇔ + + − ≥� ��

Now, we will transform the expression 1 1... ...n nn na a na a+ + − into the form of S.O.S. This can

be easily carried out via intermediary expression ( )1 11 2 ...n n

na a a− −+ +� . We have:

1 1 2( 1)( ... .. )n nn nn a a na a a− + + −

1 11 1 2 1 2 2

2

1 1 1

( 1)( ... ) ( ... ) ... ( 1) ....

( )... ... ( 2) ...

n n n n n nn n n n

i jn nn n n

i j

n a a a a a a a a n a a

a aa a na a n a a

a a

− − � �= − + + − + + + + + − −� �

−≥ + + − + −

� �

�

22

1 1 2 1 1 2 11

( ) 1 1... ... ... ... ( ... )i jn nn n n n n

i j n

a aa a na a a a a a a a a a n

a a a a

− � � � + + − ≥ = + + + ⋅ ⋅ ⋅ + −� ��

� (q.e.d)

Abstract Concreteness Method

80

ABSTRACT CONCRETENESS METHOD

I. The basis of ABC method

When using derivative to prove multi-variable inequality, there is one notable technique that is to investigate the function with one variable while the others are considered as parameters. However, if difficulties raised when we apply this technique to the symmetrical three-variable a, b, c, we should transform it into ( , , )f a b c which only contains a b c+ + , ab bc ca+ + and abc.

Then, we investigate the function with variable abc and 2 parameters a b c+ + and ab bc ca+ + follow variable abc with 2 parameters a b c+ + and ab bc ca+ + . This is the basis of ABC method. First of all, we will find the domain of abc in the following statements:

1. Clause 1. Assume , 3 3,m � � � �∈ −∞ − +∞� � � �U and real numbers a, b, c satisfying

1ab bc ca

a b c m

+ + =��

+ + =��

. Then the domain of abc is ( ) ( )2 2

2 16 2 6 2;

9 9

m X m m X m� �− + − + � �

2. Clause 2. Assume 3,m � �∈ +∞� � and nonnegative real numbers a, b, c satisfying

1ab bc ca

a b c m

+ + =��

+ + =��

then the domain of abc is ( ) ( )2 2

2 16 2 6 2Max 0, ;

9 9m X m m X m� �� − + − +

� � � � �

Proof

Consider the equation: 3 2 0X mX X abc− + − = (1)

Then two request from two clauses ⇔ Find the condition of abc such that

i) Clause 1: Equation (1) has three real roots.

ii) Clause 2: Equation (1) has three nonnegative roots.

Take ( ) 3 2f X X mX X abc= − + − � ( ) 23 2 1f X X mX′ = − +

We have: ( ) 0f X′ = ⇔2 2

1 23 3;

3 3m m m mX X− − + −= = � Variation table

i) (1) has three real roots ⇔ ( )( )

1

2

0

0

f X

f X

� ≥��

≤��

⇔ ( ) ( )2 2

2 12 6 2 6

9 9

m m X m m Xabc

− − − −≤ ≤ (2)

So the domain of abc is ( ) ( )2 2

2 16 2 6 2;

9 9

m X m m X m� �− + − + � �

X ∞− 1X 2X ∞+

ƒ′(X) + 0 – 0 +

ƒ(X)


81

ii) Notice that if 0, 0,abc a b c≥ + + ≥ 0ab bc ca+ + ≥ then , , 0a b c ≥ .

Hence (1) has three nonnegative real numbers ⇔ abc must satisfy (2) and abc ≥ 0.

⇔ ( ) ( )2 2

2 12 6 2 6Max 0,

9 9

m m X m m Xabc

� �− − − −� �≤ ≤�

And the domain of abc is ( ) ( )2 2

2 16 2 6 2Max 0, ;

9 9m X m m X m� �� − + − +

� � � � �

Note: � Assume that , 3 3,a b c m � � � �+ + = ∈ −∞ − +∞� � � �U are constructed from the equality

1ab bc ca+ + = and the inequality ( ) ( )2 3 3a b c ab bc ca+ + ≥ + + = or can be explained from

the condition ƒ(x) has three real roots then ( ) 23 2 1 0f X X mX′ = − + = should have solutions,

or 2 3 0m′∆ = − ≥ ⇔ , 3 3,m � � � �∈ −∞ − +∞� � � �U

� From clause 1 we can find out a important result is that instead of using the set ( ), ,a b c with

, ,a b c ∈ ¡ to represent every expression in 3¡ satisfying 1ab bc ca+ + = , we can use the set

( ), ,a b c ab bc ca abc+ + + + with the condition

, 3 3,a b c m � � � �+ + = ∈ −∞ − +∞� � � �U and ( ) ( )2 2

2 16 2 6 2;

9 9

m X m m X mabc

� �− + − +∈ � �

� From clause 1 we can find out a important result is that instead of using the set ( ), ,a b c with

, ,a b c ∈ ¡ to represent every expression in 3+¡ satisfying 1ab bc ca+ + = , we can use the set

( ), ,a b c ab bc ca abc+ + + + with the condition

, 3 3,a b c m � � � �+ + = ∈ −∞ − +∞� � � �U , ( ) ( )2 2

2 16 2 6 2Max 0, ;

9 9m X m m X m

abc� �� − + − +

� � ∈� � �

3. Clause 3. For every set ( )0 0 0, ,a b c 3∈ ¡ there exist ( ) ( )0 0 0 0 0 0, , ; , ,x x y z z t 3∈ ¡ such that

0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

a b c x x y z z t

a b b c c a x x x y y x z z z t t z

x x y a b c z z t

+ + = + + = + +�� + + = + + = + +��

≤ ≤��

The equality occurs if and only if ( )( )( )0 0 0 0 0 0 0a b b c c a− − − = .

Proof

Firstly we will prove the clause for all real set ( )0 0 0, ,a b c satisfy 0 0 0a b c m+ + =

and 0 0 0 0 0 0 1a b b c c a+ + = . According to clause 1 we have:

( ) ( )2 22 1

0 0 0

6 2 6 2

9 9

m X m m X ms a b c S

− + − += ≤ ≤ =


82

We will consider when 0 0 0a b c get its edge then what is the shape of 0 0 0, ,a b c ?

Assume that 0 0 0a b c = s. Consider the equation ( ) 3 2 0f X X mX X s= − + − = (1)

We have: ( ) 23 2 1f X X mX′ = − + has two solutions: 2 2

1 23 3

;3 3

m m m mX X

− − + −= =

In the other hands, ( )1 0f X = , ( )1 0f X′ = then the equation ( ) 0f X = must have double roots 1X ,

we will call the double roots is 0 0,x x and another root is 0y .

Using Viet Theorem we have:

0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0

1

x x y m a b c

x x x y y x a b b c c a

x x y s a b c

+ + = = + +�� + + = = + +��

= ≤��

� When 0 0 0a b c = S, apply the same argument we can prove the existing of ( )0 0 0, ,z z t

So we have proved the clause in the case that 0 0 0a b c m+ + = and 0 0 0 0 0 0 1a b b c c a+ + = .

Using the same ideas, the reader can have a proof for the existing in the case: 0 0 0a b c m+ + =

and 0 0 0 0 0 0 1a b b c c a+ + = − . Now let assume that 0 0 0a b c M+ + = and 0 0 0 0 0 0a b b c c a N+ + = ± ,

we will prove the existing of the set ( )0 0 0, ,x x y and ( )0 0 0, ,z z t . Indeed,

consider the numbers 0 0 01 1 1( , , ) , ,

a b ca b c

N N N

� �= � ��

satisfying 1 1 1

1 1 1 1 1 1 1

Ma b c

N

a b b c c a

� + + =��

+ + = ±��

.

As the proof above ( ) ( )1 1 1 1 1 1, , , , ,x x y z z t are exist , then we choose

( ) ( )0 0 0 1 1 1, , , ,x x y N x N x N y= and ( ) ( )0 0 0 1 1 1, , , ,z z t N z N z N t= . Then:

( )

( )

( ) ( ) ( ) ( )

0 0 0 1 1 1 1 1 1 0 0 0 0 0 0

0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 1

3 3 3 3

0 0 0 1 1 1 1 1 1 0 0 0 1 1 1 1 1 1 0 0 0

( )

( )

Mx x y N x x z N z z t z z t N M a b c

N

x x x y y x N x x x y y x z z z t t z N z z z t t z N N

x x y N x x y N a b c a b c N a b c N z z t z z t

� + + = + + = + + = + + = = = + +�� + + = + + = + + = + + = ± = ±�� = ≤ = = ≤ =�

4. Clause 4. For every set ( )0 0 0, ,a b c ∈3+¡ we can find two sets ( ) ( )0 0 0 0 0 0, , ; , ,x x y z z t ∈

3+¡

or ( ) ( )0 0 0 0 00, , ; , ,x y z z t ∈3+¡ such that

Or 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

a b c x x y z z t

a b b c c a x x x y y x z z z t t z

x x y a b c z z t

+ + = + + = + +�� + + = + + = + +�� ≤ ≤�

The equality occurs if and only if ( )( )( )0 0 0 0 0 0 0a b b c c a− − − = .


83

Or 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0

0

0

a b c x y z z t

a b b c c a x y z z z t t z

a b c z z t

+ + = + + = + +��

+ + = = + +�� ≤ ≤�

The equality occurs if and only if 0 0 0 0a b c = .

Comments: Clauses 3, 4 are not trivial and they are also the roots of ABC method. From these

clauses we can deduce that every symmetric expression ( ), ,f a b c with variables , ,a b c can be

represented as ( ), ,g A B C through these three quantities: A a b c= + + ; B ab bc ca= + + ; C abc= .

Hence reduce the number of variables, we can fix ,A B and let C free. We hope that the function

g achieves global optima when C reaches its boundary. However we will not need to know

exactly when C reach its edges, we just need to know abstractly what is the shape of , ,a b c ?

This sounds like we can find optima of a one variable function without solve its derivative.

Clauses 3, 4 will let us know about that shape of a, b, c when C reach its edge.

5. Clause 5:

Every symmetric polynomial ƒ with variables , ,a b c can be represented in the form of

polynomial ϕ with variables , ,abc ab bc ca a b c+ + + + and ( ) 1deg deg3

abc fϕ ≤ .

Remarks: The readers may understand the degrees of a polynomial with variable abc such that:

Assume that ( ) ( ), ,f a b c a b c abc ab bc ca= + + + + + are a 4-degree polynomial with variables

cba ,, . However when we consider it as a polynomial with variable abc , we consider cba ++

and cabcab ++ are constant, say m and n . Then our polynomial can be rewritten:

( ) ( ), ,f a b c abc m abc n= ϕ = + which is a linear polynomial with variable abc .

Proof

Since every symmetric three variables polynomial a, b, c can be represented as sum of

basic expression ( ) ( ) ( ), , , , ,P n Q m n R m n p , we only need to prove the representation for

( ) ( ) ( ), , , , ,P n Q m n R m n p .

( )

( ),

( , , )

n n n

m n n m m n n m m n m n

m n p m p n m n p m p n m n p m p n

P n a b c

Q m n a b a b b c c b c a a c

R m n p a b c a b c b a c b a c c a b c a b

� = + +��

= + + + + +�� = + + + + +�

( 0m n p≥ ≥ ≥ )

However, notice that:

� R can be represented by Q and abc : ( ) ( ),pR abc Q m p n p= − −

� Q can be represent by P through the equality: ( ) ( ) ( )Q P m P n P m n= − +


84

Therefore we only need to prove that P can be represented by , ,abc ab bc ca a b c+ + + + .

We will prove this using induction:

For 0; 1n n= = , the clause are trivially correct.

Assume that we proved that ( )P k can be represented as a polynomial form with variables

, , , 1abc ab bc ca a b c k h+ + + + ∀ ≤ − . Notice that this also valid for ( ),Q m n and

( ), ,R m n p 0 : 1m n p m n h∀ ≥ ≥ ≥ + ≤ − . Now we need to prove the representation for ( )P h

We have: ( ) ( ) ( ) ( )1 1,1P h a b c P h Q h= + + − − −

Also: ( ) ( ) ( )1,1 2 ( 2,1,1)Q h ab bc ca P h R h− = + + − − − .

Hence: ( ) ( ) ( ) ( ) ( )1 ( ) 2 2,1,1P h a b c P h ab bc ca P h R h= + + − − + + − + − .

On the other hand, using induction supposition, all expressions ( 1), ( 2), ( 2,1,1)P h P h R h− − −

can be represented by variables cbacabcababc ++++ ,, .

This means that our clause also correct for k = h.

In conclusion, our clause has been proved completely using induction.

Property ( ) 1deg deg3

abc fϕ ≤ was deduced trivially, because the quantity abc has three

degrees with variables a, b, c. Hence when we consider abc as a first degree variable then the

degree of abc are not greater than 13

the degrees of the polynomial with variables cba ,, .

II. ABC THEOREM

Through two previous problems, we have enough backgrounds to take an adventure in ABC World, Abstract Concreteness.

Consider ( )cbacabcababcf ++++ ,, as a one variable function with variable abc on ¡ or +¡

1. First Theorem: If the function ( )cbacabcababcf ++++ ,, is a monotonous function

then it gets the maximum and minimum values on ¡ in case ( )( )( ) 0a b b c c a− − − = , and on +¡ in case ( )( )( ) 0a b b c c a− − − = or 0abc = .

2. Second Theorem: If the function ( )cbacabcababcf ++++ ,, is a convex function then

it gets the maximum values on ¡ in case ( )( )( ) 0a b b c c a− − − = , and on +¡ in case

( )( )( ) 0a b b c c a− − − = or 0abc = .

3. Third Theorem: If the function ( )cbacabcababcf ++++ ,, is a concave function then it

gets the minimum values on ¡ in case ( )( )( ) 0a b b c c a− − − = , and on +¡ in case

( )( )( ) 0a b b c c a− − − = or 0abc = .

Remark: All of three theorems can be proved using the third and forth clauses .


85

4. Proof for the Theorems:

We will prove the first theorem using Contradiction Proof Method. Consider

( )cbacabcababcf ++++ ,, as a monotonous function on ¡ with the only variable abc .

WLOG we will only prove for the case ƒ increase and get maximum value. Assume that ƒ get

the maximum point at ( )000 ,, cba where 000 ,, cba pair-wise distinct and obtain M as

maximum value. However there exists and triple ( )000 ,, tzz satisfying:

( )( ) ( )

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

, ,

, , , ,

M f a b c a b b c c a a b c

f z z t a b b c c a a b c f z z t z z z t z t z z t

= + +

< + + + + = + + + +

This contradiction is enough for us to conclude that maximum value occur only in case

( )( )( ) 0a b b c c a− − − = . Consider ( )cbacabcababcf ++++ ,, as a monotonous function on +¡ with the only variable abc .

In case ƒ increases and we must find the maximums value then it is very similar to the above proof. In case ƒ increase and we must find the minimum value. Assume that ƒ obtain the

minimum as m at the point ( )000 ,, cba where 000 ,, cba pair-wise distinct and there are no 0

value. Then at least one of two below cases happens:

( )( ) ( )

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

, ,

, , , ,

m f a b c a b b c c a a b c

f x x y a b b c c a a b c f x x y x x x y y x x x y

= + +

> + + + + = + + + +

( )( ) ( )

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0

, ,

0, , 0, ,

m f a b c a b b c c a a b c

f a b b c c a a b c f x y x x

= + +

> + + + + = +

This contradiction is enough for us to conclude that maximum value occur only in case ( )( )( ) 0a b b c c a− − − = or 0abc = .

� The proofs for the second and the third theorem are almost similar. A reminder is that convex function get maximum, concave function get minimum when variable reach the edge

5. Consequences: We can have some useful consequences from three theorems above:

5.1. First consequence: Assume that ( )abccabcabcbaf ,, ++++ is a linear function with

variable abc , then ƒ get maximum, minimum on ¡ if and only if ( )( )( ) 0a b b c c a− − − = , on +¡ if and only if ( )( )( ) 0a b b c c a− − − = or 0abc = .

5.2. Second consequence: Assume that ( )abccabcabcbaf ,, ++++ is a quadratic trinomial

with variable abc , then ƒ get maximum on ¡ if and only if ( )( )( ) 0a b b c c a− − − = , on +¡ if

and only if ( )( )( ) 0a b b c c a− − − = or 0abc = .

5.3. Third consequence: All symmetric three variables polynomial which degrees less than or

equal to 5 get maximum, minimum on ¡ if and only if ( )( )( ) 0a b b c c a− − − = , on +¡ if and

only if ( )( )( ) 0a b b c c a− − − = or 0abc = .


86

5.4. Fourth consequence: All symmetric three variables polynomial which degrees less than

or equal to 8 with the nonnegative coefficient of 222 cba in the representation form

( )cbacabcababcf ++++ ,, get maximum, on ¡ if and only if ( )( )( ) 0a b b c c a− − − = , on

+¡ if and only if ( )( )( ) 0a b b c c a− − − = or 0abc = .

Proof

5.1. First consequence: Linear polynomial ymx + is a monotonic function. Therefore

according to the first theorem, the linear function ( )abccabcabcbaf ,, ++++ ,which is

also a monotonous function is a linear polynomial get maximum, minimum on ¡ if and only

if ( )( )( ) 0a b b c c a− − − = , on +¡ if and only if ( )( )( ) 0a b b c c a− − − = or 0abc = .

5.2. 2nd consequence: Quadratic trinomial with non-negative coefficient pnxxm ++22 is a

convex function on a continuous region. Therefore according to the second theorem, the

function ( )abccabcabcbaf ,, ++++ is a quadratic trinomial with variable abc with the

non-negative coefficient also a convex function get maximum on ¡ if and only if

( )( )( ) 0a b b c c a− − − = , on +¡ if and only if ( )( )( ) 0a b b c c a− − − = or 0abc = .

5.3. 3rd consequence: According to the fifth clause, symmetric polynomial with three

variables cba ,, whose degree are less than or equal to 5 can be represented as a polynomial

( )abccabcabcbaf ,, ++++ which is 1-degree polynomial with variable abc (since

( ) 51deg deg3 3

abc fϕ ≤ = then ( )d eg 1abcϕ = ). Therefore according to the first consequence,

the polynomial gets maximum, minimum on ¡ if and only if ( )( )( ) 0a b b c c a− − − = , on +¡

if and only if ( )( )( ) 0a b b c c a− − − = or 0abc = .

5.4. 4th consequence: According to the fifth clause, symmetric polynomial with three variables cba ,, of which degree are less than or equal to 8 can be represented as a quadratic

trinomial ( )abccabcabcbaf ,, ++++ with variable abc (since ( ) 81deg deg3 3

abc fϕ ≤ =

then ( )deg 2abcϕ = ). Otherwise the coefficient of 222 cba is nonnegative then according to

the second consequence , the polynomial get maximum on ¡ if and only if

( )( )( ) 0a b b c c a− − − = , on +¡ if and only if ( )( )( ) 0a b b c c a− − − = or 0abc = .

6. Remarks: Almost inequalities can be represented as a symmetric three variables polynomial; therefore four consequences above are really useful in many cases.

7. Some addition equalities and bonus conditions

Take .,, xyzcxzyzxybzyxa =++=++= Then we have:

7.1. 2 2 2 2 2x y z a b+ + = −

7.2. 3 3 3 3 3 3x y z a ab c+ + = − +


87

7.3. 4 4 4 4 2 24 2 4x y z a a b b ac+ + = − + +

7.4. bccaabbaazyx 5555 2235555 −++−=++

7.5. 3222346666 2312966 bcabcbacabaazyx −+−++−=++

7.6. 7 7 7 7 5 3 2 4 2 3 2 2 27 14 7 7 21 7 7x y z a a b a b a c b a a bc cb ac+ + = − + + − − + +

7.7. ( ) ( ) ( ) acbzxyzxy 22222 −=++

7.8. ( ) ( ) ( ) 23333 33 cabcbzxyzxy +−=++

7.9. ( ) ( ) ( ) bccaacbbzxyzxy 22224444 424 ++−=++

7.10. ( ) ( ) ( ) 3222235555 5555 accbbcacabbzxyzxy −++−=++

7.11. ( ) ( ) ( ) cabxzzxzyyzyxxy 3−=+++++

7.12. ( ) ( ) ( ) acbbaxzzxzyyzyxxy −−=+++++ 22222222 2

7.13. ( ) bccaabbaxzzxzyyzyxxy 53)()( 223333333 +−−=+++++

7.14. ( ) ( ) ( ) bccaabxzxzzyzyyxyx −−=+++++ 22222222 2

7.15. ( ) ( ) ( ) 2223333333 53 cbacbcaabxzxzzyzyyxyx −+−=+++++

7.16. ( )( ) ( ) 332222222 69 bcabaczxyzxyxzzyyx +−+=++++

7.17. ( )( ) ( )3 3 3 3 3 3 2 2 5 3 2 47 5x y y z z x xy yz zx a c a a b ab c b+ + + + = + − + +

7.18. ( )( )3 2 3 2 3 2 2 3 2 3 2 3 4 2 2 2 2 2 5 37 5x y y z z x x y y z z x a c a bc b c b b ac+ + + + = + + + −

7.19. ( )( )4 4 4 4 4 4

5 3 2 2 2 2 7 5 3 2 4 212 16 13 7 14 7

x y y z z x xy yz zx

b ab c a bc b c ca ca b ca b a c

+ + + + =

= − − + + − + +

7.20. The 3-degree equation 3 2 0u au bu c− + − = has real roots zyx ,,

⇔ ( )2 3 2 2 327 18 4 4 0c ab a c a b b− + − + − ≥ (1)

7.21. The 3-degree equation 3 2 0u au bu c− + − = has positive real roots zyx ,, > 0

⇔ ( )2 3 2 2 327 18 4 4 0

0, 0, 0

c ab a c a b b

a b c

�− + − + − ≥�� > > >�

7.22. The 3-degree equation 3 2 0u au bu c− + − = has roots zyx ,, which are length of sides of

a triangle ⇔

( )2 3 2 2 3

3

27 18 4 4 0

4 8 0.

0, 0, 0

c ab a c a b b

a ab c

a b c

�− + − + − ≥�� − + >�� > > >�


88

Now we will go through some specifies examples:

Problem 1 [Nguyen Anh Cuong] Give cba ,, > 0. Prove that:

1. 3 3 3 2 2 2

23

abc ab ac bca b c a b c

+ ++ ≥+ + + +

(1) 2.23 3 3 2 2 21

4 4a b c a b c

abc ab ac bc

� �+ + + ++ ≥ � �+ +� � (2)

Solution

1. (1) ⇔ ( ) ( ) ( ) ( )2 2 2 3 3 3 3 3 32 03

P abc a b c a b c a b c ab bc ca= + + + + + − + + + + ≥

P has third degrees so it will get the minimum values when ( )( )( ) 0a b b c c a− − − = or 0abc = .

� In case ( )( )( ) 0a b b c c a− − − = , assume that ca = , the inequality is equivalent to:

( )( )

( ) ( )2 2

2 4

3 3 2 2 2 2 3 3

2 2 1 20 0.

32 2 2 3 2

a b a ab a ba b a b a b

a b a b a b a b

+ +� �+ ≥ ⇔ − − ≥ ⇔ − + ≥ + + + +� �

� In case 0abc = , assume that 0c = , the inequality is equivalent to:

( )22 22 2

23 0.

3ab

a b a ba b

≥ ⇔ + + − ≥+

2. Notice that we can transform (2) to a 7-degree symmetric polynomial with variables a, b, c but it is only a 1-degree polynomial with variable abc , therefore according to the 1st consequence, we just need to consider two cases:

� In the case ( )( )( ) 0a b b c c a− − − = , assume that ca = the inequality is equivalent to: 2 23 3 2 2 3 3 2 2

2 2 2 2

1 32 2 2 2 14 44 2 4 2

a b a b a b a ba b a ab a b a ab

� �� + + + ++ ≥ ⇔ − ≥ −� � � � � �+ +� � � � � �

( ) ( ) ( ) ( )( )

( ) ( )2 2 2 2

2 2 22 22

2 3 2(2 ) 0

4 2

a b a b a b a b aba b b a a

a b a ab

− + − + + � �⇔ ≥ ⇔ − − + ≥� �+

� In the case 0abc = , the inequality is trivially correct.

Problem 2. Give cba ,, > 0. Prove that: ( )( ) ( ) ( )2 2 2

91 1 14

ab bc caa b b c c a

� �+ + + + ≥ + + +� �

(1)

(Iran Olympiad 1996)

Solution

Transform (1) to a 6-degrees symmetric polynomial with a, b, c and a 2-degrees polynomial with variable abc and positive coefficient:

( )( )( )[ ] ( ) ( ) ( ) ( ) ( ) ( ) ( )

( )

2 2 2 2 2 2 2

2

( , , )

9 4 (*)

( 0) 0

f a b c ab bc ca abc

a b b c c a ab bc ca a b b c b c c a c a a b

m abc nabc p m

+ + + + =

� �+ + + − + + + + + + + + + +� �

= + + ≥ ≤

in which , ,m n p are quantities containing constant or ,a b c ab bc ca+ + + + which we also consider

as constant.


89

We can give more detail for this as following: ( )( )( )a b b c c a+ + + has the form mabc n+ so

[ ]29 ( )( )( )a b b c c a+ + + has the form ( ) 22m abc nabc p+ + .

( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2 24 4ab bc ca a b b c b c c a c a a b mA� �+ + + + + + + + + + =� � where m ab bc ca= + +

are considered as constant, A is a 4 degrees polynomial so its form is also mabc n+ .

Therefore the expression (*) has the mentioned form.

Then the LHS function gets the maximum value only in the case ( )( )( ) 0a b b c c a− − − = or 0abc =

� In the case ( )( )( ) 0a b b c c a− − − = , assume that ca = the inequality is equivalent to

( )( )

( )( ) ( )

( )2 222 2 2 2

1 2 9 2 12 0 0

44 2

a ba ab a b b a b

a a b a a b a b

+� � � �+ + ≥ ⇔ − − ≥ ⇔ − ≥� � + + +� � � �

� In the case 0abc = , assume that c = 0, the inequality is equivalent to

( )( )

( )( ) ( )2 2 2 2

2 2 2 2

1 1 1 9 1 10 4 4 7 0

4 4ab a b a b a b ab

aba ba b a b

� � � �+ + ≥ ⇔ − − ≥ ⇔ − + + ≥� � � �+ +� � � �

Following is some another example that we can not use ABC Theorem or its consequences

directly until we have some algebraic transformation.

Problem 3. Given , ,a b c > 0 satisfying 2 2 2 1a b c+ + = . Prove that:

3 3 33

a b ca bc b ac c ab

+ + ≥+ + +

(1)

(Russia Olympiad 2005)

Solution

Comments: If we transform (1) to a symmetric polynomial then we will obtain a nine degrees

polynomial with variable a, b, c; and third degrees with variable abc. These cases are not in

our consequences.

Therefore we should have some algebraic transformation before think of ABC.

Take ; ;bc ac ab

x y za b c

= = = � 1xy yz xz+ + = and (1) ⇔ 1 1 1

3xy z yz x zx y

+ + ≥+ + +

(2)

Transform (2) to a second degrees polynomial with variable xyz with the coefficient for 2xyz

is non-negative. We just need to consider two cases as usual.

Case 1: x z= . Then (2) ⇔ 2

2 13

xy x x y+ ≥

+ + then 22 1.xy x+ =

Replace y by 21

2xx

− then we must prove that: [ ]2 2

2

2 1 3, 0;11 1

2 2

xx xx x

x

+ ≥ ∀ ∈− −+ +

.

This is no longer too difficult with Fermat Theorem.


90

Case 2: z = 0. Then (2) ⇔ 1 1 1

3xy x y

+ + ≥ where 1xy = .

We have 1 1 2 1 1 1 1 2

3x y xy x y xyxy xy

+ ≥ � + + ≥ + = .

Problem 4 [Nguyen Anh Cuong] Give positive real numbers zyx ,, . Prove that

( )4 4 4

2 2 2 2 2 2 2 2 2

21 2

xy yz xzx y zx y y z z x x y z

+ ++ + + ≥ ++ + + +

(1)

Solution

For this problem, thinking about transform it to a polynomial is really a problem.

In this case we should forget about the consequences and apply theorems for it.

As usual denote that , ,a x y z b xy yz zx c xyz= + + = + + = , use the equality mentioned on

previous part to transform our inequality to : 4 2 2

2 2

2 2 4 21 2

2 2a a b b ac b

b ac a b

− + + + ≥ +− −

The function with variable c is simple a monotonous one. According to the first theorem, its

minimum occurs only in cases ( )( )( ) 0a b b c c a− − − = or 0abc = .

Case 1: zx = . Then (1) ⇔ ( )24 4

4 2 2 2 2

2 221 2

2 2

x xyx y

x x y x y

++ + ≥ ++ +

(2)

We can assume that 1=x because of its homogeneous property, then

(2) ⇔ ( ) ( )

( ) ( )( )

( ) ( )

2 224

2 2 2 4 2 2 2

2 2 1 1 2 121 2

2 1 2 2 1 2 2 1 2 2 2 1

y y yy

y y y y y y y y

+ − −+ − ≥ − ⇔ ≥+ + + + + + + + + +

Notice some evaluation:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

4 2 3 2 4 2 2 3

2 2 2 2

2 2 1 2 2 2 1 2 2 1 2 1 2 2

2 2 1 2 2 2 2 1 2 2

y y y y y y y y

y y y y y y y y

+ + ≤ + � + + + + ≤ + + +

+ + ≥ + � + + + + ≥ + + +.

So it suffices to prove that:

( ) ( ) ( )2

4 3 23 2 2

1 25 2 5 2 2 0

2 2 2 1 2 2

yy y y y

y y y y

+≥ ⇔ + + − + + ≥

+ + + + + +

Case 2: 0z = . Then (1) ⇔ 4 4

2 2 2 2

21 2

x y xyx y x y

+ + ≥ ++

(2)


91

We have: ( ) 22 2

4 4 2 222

x yx y x y

++ ≥ ≥ �

4 4 2 2

2 21

22x y x y

xyx y+ +≥ ≥ , therefore:

( ) ( ) ( )4 4 2 2 2 2

42 2 2 2

22 2 1 2 1 2 1 2

2 22x y x y xy x y

LHSxy xyx y x y

� �+ + +≥ − + + ≥ − + ⋅ ≥ +� �� +� �

Problem 5. Given , ,a b c ∈ ¡ and 2 2 2 2a b c+ + = . Prove that: 3 3 3 2 2a b c abc+ + − ≤

(Mongolia MO 1991)

Proof

The expression given is not a normal polynomial. We have two choices in this case; prove two inequalities or take its square. In this part I will square it since it is still valid for ABC Theorem. Indeed the polynomial obtained after squaring is a six degrees polynomial with

positive coefficient for 222 cba and we must find its maximum. According to ABC, we refer to two cases:

Case 1: ( )( )( ) 0a b b c c a− − − =

Given x, y: 2 22 2x y+ = . Prove that: 3 3 22 2 2x y xy+ − ≤ (1)

We have (1) ⇔ ( ) ( )2 33 3 2 2 22 2x y xy x y+ − ≤ + (2)

If 0y = then we have the equality.

When 0y ≠ , divide two sides for 6y and replace xy

by t , then (2) ⇔ ( ) ( )2 33 22 2t t t− + ≤ +

We have: ( ) ( ) ( )2 2 33 3 6 3 6 4 2 22 2 4 2 6 12 8 2t t t t t t t t t− + ≤ + = + + < + + + = +

Case 2: 0abc =

Give x, y: 2 2 2x y+ = . Prove that: 3 3 2 2x y+ ≤

Indeed, we have: ( ) ( ) ( )33 3 4 4 2 2 2 2 2 2x y x y x y x y+ ≤ + + ≤ + =

Problem 6 [Nguyen Anh Cuong]

Given non-negative numbers cba ,, satisfying 2a b c+ + = . Prove that:

2 2 2 2 2 20 2a b b c c a ab bc ca≤ + + + + + ≤

Solution

These are no longer polynomial, and to transform it to polynomial is really a problem.

There are no better way in mind, square it and see:

( ) ( )2 2 2 2 2 2 2 2 2 2 2 22 4a b b c c a ab bc ca a b b c c a ab bc ca+ + + + + + + + + + ≤ (1)


92

Still not a polynomial, we can transform to a polynomial in the next square step but our

polynomial has too big degrees already.

Try to transform it with 2, ,A a b c B ab bc ca C abc= + + = = + + = :

( ) 2 3 2 31 2 3 2 9 (8 12 ) 4 2 9 (8 12 ) 4 2 3B C C B C B C B C B B C⇔ − + + − + ≤ ⇔ + − + ≤ − +

( )( ) ( ) 22 3 2 2

2 3 2

4 9 8 12 4 2 3 9 6(4 2 ) (4 2 )

27 (8 36 ) 4 (4 2 ) 0

C B C B B C C B C B

C B C B B

⇔ + − + ≤ − + = + − + −

⇔ + − + − − ≤

Not it is very nice to apply ABC already, the function with variable C is convex and we are

finding its maximum value. As usual consider two cases:

Case 1: ( )( )( ) 0a b b c c a− − − =

Given , 0x y ≥ satisfying 2 2x y+ = . Prove that 3 2 22 2x x y xy+ + ≤

( ) ( ) [ ]

( ) ( ) [ ] [ ]

23 2

2 2

2 2 2 2 1, 0;1

1 3 3 1 0 , 0;1 3 3 1 0, 0;1

x x x x x x

x x x x x x x

⇔ + − + − ≤ ∀ ∈

⇔ − − + ≤ ∀ ∈ ⇔ − + ≥ ∀ ∈

Case 2: 0abc =

Given , 0x y ≥ satisfying: 2x y+ = . Prove that: 2 2 2x y xy+ ≤

We have: ( ) ( )2 2 2 22

x yx y xy xy x y x y

++ = + ≤ + =

Proposed Problem

Problem 7. Give non-negative numbers a, b, c satisfying: 1=++ zxyzxy .

Prove that: 924

xy yz zx xyz+ + ≤ +

Problem 8. Give positive numbers cba ,, . Prove that:

2 2 2 2 2 261 1 1

2 2 2a bc b ca c ab a b c ab bc ca+ + ≥

+ + + + + + + +

Problem 9. [Darij Grinberg- Old and New Inequality]

Give cba ,, as positive numbers. Prove that:

( ) ( ) ( ) ( )2 2 2

94

a b ca b cb c c a a b

+ + ≥+ ++ + +


93

Problem 10. [Mircea Lascu – Old and New Inequality]

Give cba ,, as positive numbers. Prove that:

( )4b c c a a b a b ca b c b c c a a b+ + ++ + ≥ + +

+ + +

Problem 11. [Vietnam TST, 1996]

Prove this inequality for all real numbers cba ,,

( ) ( ) ( ) ( )4 4 4 4 4 447

a b b c c a a b c+ + + + + ≥ + +

Problem 12. [Nguyen Anh Cuong] Give positive cba ,, . Prove that

2 2 22 6 2

a b b c c a ab bc cac a b a b c+ + + + ++ + + ≥ +

+ +

Problem 13. [Russia MO] Assume that zyx ,, are positive numbers of which sum is 3.

Prove that: x y z xy yz zx+ + ≥ + +

Problem 14. [Vasile Cirtoaje] Give a, b, c ≥ 0 and 2 2 2 3a b c+ + = . Prove that:

( ) ( ) ( )2 2 2 1ab bc ca− − − ≥

Problem 15. [Nguyen Anh Cuong]

Give non-negative numbers , ,x y z : 2x y z+ + = . Prove that:

3 3 3 3 3 30 2x y y z z x xy yz zx≤ + + + + + ≤

Problem 16. [Nguyen Anh Cuong] Give , , 0a b c ≥ and 2 2 2 1a b c+ + = .

Prove that: 3 3 3

52

bc ca aba a b b c c

+ + ≥− − −


94

III. ABC EXTRA

As you have seen in previous part, almost the evaluation of ABC based on abc . Therefore if

the problems come with a condition with abc then it is really a problem with ABC. We will

consider some problems having this condition and try to solve it with ABC.

A trick we might use is that we can destroy the condition in many ways and refer to a

homogeneous inequality. And then we can think of using ABC , specified in this problems.

Problem 1 (Vasile Cirtoaje, MS, 2006)

Given positive numbers cba ,, satisfying 1abc = . Prove that:

( )2 2 2 3 1 1 162

a b c a b ca b c

+ + + ≥ + + + + + (1)

Proof

The inequality (1) ⇔ ( )2 2 2 362

a b c a b c ab bc ca+ + + ≥ + + + + +

( ) ( ) ( )22 12 3 7a b c a b c ab bc ca⇔ + + + ≥ + + + + + (2)

There exits positive numbers zyx ,, such that 2 2 2

, ,x y z

a b cyz zx xy

= = = .

Then (2) ⇔ ( )2 3 3 33 3 3 3 3 3 3 3 3

2 2 2

32 12 7

x y zx y z x y y z z xxyz xyz x y z

� � + ++ + + ++ ≥ + ⋅� ��

( ) ( ) ( )23 3 3 2 2 2 3 3 3 3 3 3 3 3 32 12 3 7x y z x y z x y z xyz x y y z z x⇔ + + + ≥ + + + + + (3)

Because of homogeneous property (3), we can assume that 1, ,x y z xy yz zx v xyz w+ + = + + = =

We have: 3 3 3 3 3 3 3 3 3 3 21 3 3 ; 3 3x y z v w x y y z z x v vw w+ + = − + + + = − +

Then (3) ⇔ ( ) ( ) ( )2 2 3 22 1 3 3 12 3 1 3 3 7 3 3v w w v w w v vw w− + + ≥ − + + − +

( ) ( ) ( )2 32 1 3 9 1 3 7 3 0v v w v vw⇔ − + − + − ≥

The first degrees polynomial with variable w , good condition for ABC - Theorem for (3).

Case 1: 0, 1z y= =

The inequality (3) is equivalent to: ( ) 23 32 1 7x x+ ≥ ⇔ ( ) 23 32 1 0x x− + ≥

Case 2: 1== zy

The inequality (3) is equivalent to: ( ) ( ) ( )23 2 3 32 2 12 3 2 7 2 1x x x x x+ + ≥ + + +

( ) ( ) 26 4 3 2 4 3 22 3 6 12 6 1 0 2 4 3 4 1 1 0x x x x x x x x x x⇔ − − + − + ≥ ⇔ + + − + − ≥


95

Problem 2. Given cba ,, ≥ 0 satisfying: 4ab bc ca abc+ + + = .

Prove that: 1 1 1 a b ca b c

+ + ≥ + + (1)

Proof

There exits zyx ,, such that: 22 2

, ,yx z

a b cy z x z x y

= = =+ + +

Then (1) becomes: 4x y y z yx z x z

z y x y z x z x y� �+ +++ + ≥ + +� �+ + +� �

Because of homogeneous property we can assume that 1, ,x y z xy yz zx v xyz w+ + = + + = =

and rewrite inequality as: 11 1

41 1 1

y yz x x zz y x x y z

−− − � �+ + ≥ + +� �− − −� �

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

1 1 1 1 1 1

4 1 1 1 1 1 1

xy z xz y yz x x y z

xyz x y z y z x z x y

� � � �⇔ − + − + − − − −� ��

� �≥ − − + − − + − −� �

( ) ( ) ( ) ( )2 23 4 1 2 3 9 4 1 0v w v w w v w w v w v⇔ − − ≥ − + ⇔ + − − ≤

We can apply ABC now, consider two cases:

In case z = 0, the inequality is trivially correct.

In case 1y z= = (we can consider them as one because of homogeneous property):

( ) ( ) ( ) ( ) ( )22 22 1 4 2 1 2 1 8 2 1 02 1xx x x x x x

x x+ + ≥ + ⇔ + + + ≥ ⇔ − ≥

+

Problem 3. Given a, b, c satisfying 1, 1, 1a b c≥ ≥ ≥ and 2a b c abc+ + + = .

Prove that: ( )1 1 13 6a b ca b c

+ + + ≥ + + (1)

Proof

There exits positive numbers zyx ,, such that , ,y z x yx z

a b cx y z+ ++= = =

where zyx ,, are length of sides of a triangle.

(1) ⇔ ( ) 1 1 1 6yx zx y z

x y z y z z x x y� �� + + + + ≥ + +� � � �+ + +� � � �

(2)

Because of homogeneous property, we can assume that 1, ,x y z xy yz zx n xyz p+ + = + + = =

(2) ⇔ ( ) 61 1 1

xy yz zx yx zx y z

xyz x y z+ + � �+ + ≥ + +� �− − −� �


96

( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )

1 1 1 1 1 16

1 1 1

x y z y z x z x yxy yz zxx y z

xyz x y z

− − + − − + − −+ +⇔ + + ≥ ⋅− − −

( ) ( )( )

2 36

1

x y z xy yz zx xyzxy yz zxx y z

xyz xyz x y z xy yz zx

+ + − + + ++ +⇔ + + ≥ ⋅

− − + + + + +

( ) ( )2 2 26(1 2 3 )6(1 2 ) 3 3 6 11 0

n pnn n p n p p p n p n

p n p− +⇔ ≥ ⇔ − ≥ − + ⇔ + − − ≤

−

Since , ,x y z are length of sides of a triangle then we can take vuzwuywvx +=+=+= ,, .

We have: 22( ), ( )a b c u v w xy yz zx uv vw wu u v w+ + = + + + + = + + + + + and

( )( )xyz u v w uv vw wu uvw= + + + + − then the expression (1) with variables wvu ,, still be a

2-degree polynomial with variable uvw and positive coefficient for ( ) 2uvw and we must find

maximum value. We can apply ABC now, consider two cases:

Case 1: 0w = � yxz += . We can assume .1=y

Inequality (2) becomes: ( ) ( )1 1 12( 1) 1 6 11 2 2 1

xxx x x x

+ + + ≥ + ++ + +

( ) 22 2 212 3 6 1( 2)(2 1)

x xxx x x

� �+ +⇔ + + ≥ + + +� �

( )2 2

2

1 3 12 2 5 2

x x xx x x+ + +⇔ ≥

+ +

( ) ( ) ( )2 2

2 22

1 1( 1) 2 3 2 0

2 2 5 2x x

x x xx x x

− −⇔ ≥ ⇔ − + + ≥+ +

(is always true) � (q.e.d)

Case 2: zywv =�= . We can assume 2,1 ≤== xzy

Inequality (2) becomes ⇔ ( ) ( ) ( )1 22 2 62 1xx

x x+ + ≥ +

+

2( 1)( 2)(2 1) 3 ( 4)x x x x x x⇔ + + + ≥ + + ( ) ( ) 22 1 0x x⇔ − − ≥ (is always true) � (q.e.d)

The problem was solved.

However in many cases, we can not destroy condition and refer the old problem to a

homogeneous problem, and ABC extra was born for these problems.

ABC – EXTRA

i) Given cba ,, are all real numbers or positive real numbers. If we fix ,abc a b c+ +

(consider it as constant) then ab bc ca+ + have its maximum value or minimum value only

when ( )( )( ) 0a b b c c a− − − = .

ii) Given cba ,, are all positive real numbers. If we fix ,abc a b c+ + (consider it as constant)

then a b c+ + have its minimum value only when ( )( )( ) 0a b b c c a− − − = .


97

Proof

i) Assume that 1=++ cba and mabc = (the case ncba =++ can be refer to this cases through a algebraic transformation)

We will prove that cabcab ++ can obtain maximum and minimum value only when

( )( )( ) 0a b b c c a− − − = in both cases , ,a b c ∈ ¡ and , ,a b c +∈ ¡ .

Assume that Scabcab =++ . Again we refer to the third degrees equation which have cba ,, as its roots:

Take mSXXXXf −+−= 23)( . We have: ( ) 23 2f X X X S′ = − + .

Equation ( ) 0f X′ = has two solutions 1 21 1 3 1 1 3

;3 3

S SX X

+ − − −= =

The equation has three roots if and only if ( ) 02 ≥Xf , ( ) 01 ≤Xf .

We have:

( ) ( ) 09260 22 ≥−+−⇔≥ mSXSXf , assume that its solution range is 2XR

( ) ( ) 09260 11 ≤−+−⇔≤ mSXSXf , assume that its solution range is 1XR

Firstly we prove our theorem for the case , ,a b c ∈ ¡ .

Assume that maxmin , SS are minimum and maximum values in the set 1 2X XR RI (the union of

these two ranges are not null or ( ) 0f X = has no solution for all values of S ). Notice that

maxmin , SS are two solution of ( ) 0926 1 =−+− mSXS or ( ) 0926 2 =−+− msXS ( these

value surely exits or ( ) 0f X = has solutions when S reaches infinity values), then we have

.maxmin SSS ≤≤ Now let consider cba ,, when S reaches these two values .

The first point we can see is that ( ) 0f X = still have three solutions since 1 2min max, X XS S R R∈ I .

The second point, because maxmin , SS are solutions of ( ) 16 2 9 0S X S m− + − = or

( ) 26 2 9 0S X s m− + − = , therefore ( ) 01 =Xf , or ( ) 02 =Xf . In this case the equation

( ) 0f X = has double root, or in the other hands, the shape of ( )cba ,, is ( )yxx ,, .

Now we prove our theorem for the case , ,a b c +∈ ¡ . At first we must have 0m ≥ . Notice that

210 XX RR ∩∉ because there are no real number cba ,, such that

0,0,1 ≥=++=++ abccabcabcba . This means we can separate 21 XX RR ∩ in smalls

[ ],i ix y such chat 0i ix y > . Let 3R are the set of 21 XX RR ∩ ignored the negative ranges. And

let maxmin , SS are minimum and maximum values in the set 3R . We have: maxmin SSS ≤≤ .

Using the same argument as above we also can obtain the result that S reach its edge if and

only if ( )( )( ) 0a b b c c a− − − = .


98

ii) The readers can ready use the same arguments to prove the ABC Extra Theorem 2.

However in the case for a b c+ + , we can not limit the upper bound for a b c+ + when a, b, c ≥ 0.

Therefore in this case we can only have minS . And we will come out with the result cba ++

get minimum value if and only if ( )( )( ) 0a b b c c a− − − = .

Let see some example to know how this theorem can help us in a lot of cases:

Problem 4 [Hojoo Lee] Give positive numbers cba ,, satisfying 1≥abc .

Prove that: 1 1 1

11 1 1a b b c c a

≥ + ++ + + + + +

(1)

Proof

Notice that the LHS function decrease when a increase. Therefore it is enough to prove the inequality when 1=abc (for the case 1≥= kabc , we have:

1 1 1 1 1 11

1 1 1 11 1a aa b b c c a b cb ck k

+ + ≤ + + ≤+ + + + + + + ++ + + +

)

The inequality (1) ⇔

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( , , ) 1 1 1 1 1 1 1 1 1 0f a b c a b b c c a a b b c b c c a c a a b= + + + + + + − + + + + − + + + + − + + + + ≥

Notice that the degrees of cabcab ++ in ( )cbaf ,, in only one ( ( )cbaf ,, is only a third

degrees polynomial with variables cba ,, ). Therefore when we fix cba ++ then ( )cbaf ,, has

its minimum value when cabcab ++ has its minimum value, then ( )( )( ) 0a b b c c a− − − = . We

refer to this problem:

Given yx, > 0 satisfying 12 =yx . Prove that: 1 2

11 2 1x x y

+ ≤+ + +

(2)

Replace y with 2

1x

into (2) we have:

2

1 21

11 2 1x xx

+ ≤+ + +

( ) ( )

( ) ( )2

3 2

2 1 10

1 2 1

x x x

x x x

− +⇔ − ≤+ + +

Hence the proof is completed.

Problem 5 [Bui Viet Anh] Give three positive numbers cba ,, . Prove that:

2 2( )( )( )

abca b cb c c a a b a b b c c a

+ + + ≥+ + + + + +

(1)

Solution

For this problem, even when we fix ,abc ab bc ca+ + or a b c+ + we can not obtain a nice

function (means monotony, concave or convex). However we can see some relation between

a b cb c c a a b

+ ++ + +

and ( )( )( )

abca b b c c a+ + +

, they are sum and product of , ,a b cb c c a a b+ + +

.


99

Therefore we take , ,a b c

x y zb c c a b a

= = =+ + +

, the problem is that how to connect zyx ,, .

Here I will give a condition for zyx ,, , and with this condition we can convert zyx ,, to

, ,a b c again. And then we will come out with the equivalent problem:

Give 0,, ≥zyx satisfying: 1 1 1

2 2 11 1 1

xyz xy yz zxx y z

+ + = ⇔ + + + =+ + +

.

Prove that: 2 2x y z xyz+ + + ≥ .

So we will fix xyz and zxyzxy ++ and refer to the problem:

Give 0, ≥ba and 2 22 2 1a b a ab+ + = . Find the minimum of: 2 2a b a b+ + .

Replace ( )2

2

1 11

22 2a a

b aaa a

− −= = ≤+

, we need to prove ( )12 2 1 22

aa a aa

−+ + − ≥ (2)

Indeed, the inequality (2) ⇔ ( ) ( )1 2 1 21 2 1 2

2 1 2 1a a a

a aa a a a

− + − ≥ − ⇔ + ≥− −

In the other hands 2 2 2

2 211 (1 )

2

a a aa

a aa a a= ≥ ≥

− +− −.

Therefore: 41 2 1 12 2 2 2 2 2 2 2

2 1 2 2a

a aa a a a

+ ≥ + ≥ = ⋅ ≥−

.

The proof is end here.

Problem 6. Give positive numbers zyx ,, . Prove that

2 2 2

3 3 3 3 3 3 3 3 3

2 313

x y z xyz

x y y z z x x y z+ ≥

+ + + + (1)

Proof

Take 2 2 2

, ,yz xyxz

a b cx y z

= = = , the inequality (1) ⇔ 2 1 3

3a b c ab bc ca+ ≥

+ + + +

where cba ,, are positive numbers satisfying 1=abc .

After this algebraic transformation, the problem is quite trivial with extraABC − .

We refer to the only case:

Give positive numbers 1:, 2 =yxyx . Prove that: 2

2 1 32 3 2x y x xy

+ ≥+ +


100

2

3 322

2 2 3 2 2 31 1

1 23 32 1 22

x xx xx x

xx

⇔ − ≥ − ⇔ − ≥ −+ ++ +

( ) ( ) ( ) ( )( )

2 2

3 3

2 1 2 1 12 3 2 1

x x x xx x

+ − + −⇔ ≥+ +

( ) ( )24 34 10 2 1 0x x x x⇔ + − + − ≥

The readers might try this problems with ABC Theorem− , and you might see the efficient of extraABC − .

Proposed Problems

Problem 7. Given cba ,, > 0. Prove that: ( )3 23 3

x y zxyz xy yz zx

+ + + ≥ + +

Problem 8. Give positive numbers cba ,, of which product is 1. Prove that

2 2 2 3a b c a b c ab bc ca+ + + ≥ + + + + +

Problem 9. Prove this inequality for non-negative numbers cba ,, :

( )2 2 2 2 1 2a b c abc ab bc ca+ + + + ≥ + +

Problem 10. Give positive numbers cba ,, of which product is 1.

Prove that 2 2 2

1 1 1a b c

a b c+ + ≥ + +

+ + +

Problem 11. [Vietnam MO 1996]

Given cba ,, ≥ 0 satisfying 4a b c abc+ + + = . Prove that a b c ab bc ca+ + ≥ + +

Problem 12 [Le Trung Kien, Vo Quoc Ba Can]

Given 0,, ≥cba satisfying 96 =+++ abccabcab . Prove that: 63 ≥+++ abccba

Problem 13. Give positive numbers zyx ,, satisfying: 1≥xyz . Prove that

1 1 11

2 2 2x y z+ + ≤

+ + +

Problem 14. Give positive numbers cba ,, of which product is 1. Prove that

( ) ( ) ( ) ( )66 2 2 23 1 1 1 8a b c a b c+ + + ≤ + +


Give positive numbers cba ,, satisfying: ( )( )( ) 8a b b c c a+ + + = .

Find the maximum and minimum values of: P abc a b c= + + +

Problem 16. Given cba ,, satisfy 1 1 1 3a b c

+ + = . Prove that ( )93 14

a b c abc+ + − ≤ −


101

IV. ABC – OPEN THE WINDOW

In this part, we will again consider a class of problem that the old ABC can not handle. That

is the class in which variables are bounded in close sets. For these problems, there are some

classic ideas to solve that I will introduce again in this part, and later is a new convention in

ABC to handle this close set

Problem: Give [ ]1 2, ,..., ,nx x x a b∈ . Prove that ( )1 2, ,..., nf x x x C≥

Where Cba ,, are constant given.

In these cases, the equality of inequality usually occurs when variables reach theirs edge,

means a or b . In these cases, we will try to prove that

( ) ( ) ( ){ }1 2 2 2, ,..., min , ,..., , , ,...,n n nf x x x f a x x f b x x≥ . At last we will obtain the minimum value

for our function, occurring when some variables are equal to a , and the rest get b as their

values . We will consider some example to make this idea clearer:

Problem 1. Give [ ], , 0,1a b c ∈ . Find the maximum value of: ( ), ,f a b c a b c ab bc ca= + + − − −

Solution

We will prove that ( ) ( ) ( ){ }, , max 0, , , 1, ,f a b c f b c f b c≤ (*).

Following is a direct proof using algebraic only:

( ) ( ) ( ) ( ) ( )( ) 2, , 0, , , , 1, , 1 1 0f a b c f b c f a b c f b c c c a b� � � �− − = − − − ≤� �� , from here we have (*)

Or using a bit more advance view, we can find out the result faster with a notice that

( ), ,f a b c is a monotonic function when we only consider variable a . Then we can find out

(*) immediately.

Apply the same thing for the rest variables cb, , we will find out this property :

( ) ( ) ( ) ( ){ }( , , ) max 0,0,0 ; 0,0,1 ; 0,1,1 ; 1,1,1f a b c f f f f≤ .

From here we can conclude that the maximum value of ƒ(a, b, c) is 1 .

The inequality occurs , for example, when: 0, 1a b c= = = .


Given [ ], , 1, 2a b c ∈ . Find the maximum value of a b c

b c c a a b+ +

+ + +

Solution

We will consider cb, as constant, and a is variable: ( )( ) ( )3 3

2 2, , 0

b cg a b c

c a b a′′ = + >

+ +


102

Therefore g is a convex with variable a on +¡ , i.e. g will get its maximum value when a

reaches its edges.

From here we can deduce what we hope: ( ) ( ) ( ){ }, , max 1, , , 2, , .g a b c g b c g b c≤

Apply the same thing for the rest variables cb, , we will find out a property:

( ) ( ) ( ) ( ) ( ){ }, , max 1,1,1 ; 2,1,1 ; 2;2;1 ; 2, 2, 2g a b c g g g g≤ .

From here we can conclude that the maximum value of g is 1912

, occurs when ( ) ( ), , 2,2,1a b c =

Through two examples above, the readers may have some familiar feeling when meet the kind of

problems again. Now I will represent another method to kill these problems, an ABC approach.

In above parts, we see that ABC can only apply for problems when domain is ¡ or +¡ , so

to apply ABC , we must resize the given domain to ¡ or +¡ .

Problem 1. Given [ ], , 0,1a b c ∈ . Find the maximum value of: ( ), ,f a b c a b c ab bc ca= + + − − −

Solution

Again we meet the problem in section one, now we need to [ ]0;1 to +¡ .

Notice that if we transform 1au

→ , we will obtain the domain [ ]1; +∞ for u . It is still not +¡ ,

we continue the next step , map 1u x→ + and we will obtain the domain [ ]0, +∞ for x . So we

have transform 11

ax

→ + to pull [ ] [ ]0;1 0;→ +∞ . From these ideas we can come out with a solution:

Take 1 1 1

, ,1 1 1

a b cx y z

= = =+ + +

where , ,x y z +∈ ¡ . The expression given becomes:

( ) ( ) 1 1 1 1 1 1, , , ,

1 1 1 (1 )(1 ) (1 )(1 ) (1 )(1 )f a b c g x y z

x y z x y y z z x= = + + − − −

+ + + + + + + + +

( ) ( ) ( ) ( ) ( ) ( )1 1 1 1 1 1 3(1 )(1 )(1 ) 1

y z z x x y x y z xy yz zx x y zx y z xyz xy yz zx x y z

+ + + + + + + + − − − − + + + + += =+ + + + + + + + + +

Now we can easily see that ( ), , 1g x y z ≤ , no need to apply ABC anymore, of if we want we

still can apply ABC to refer to two variables problems.

There are two advantages for us when pull a small domain to +¡ like this. First thing is that

we can ABC theorem as above problem; second thing is that we can evaluate the inequality’s

variable with 0 which is always easier.


103

Problem 3 Given [ ], , 1, 2x y z ∈ . Prove that: ( ) 1 1 1( , , ) 10g x y z x y zx y z

� �= + + + + ≤� ��

Again we use the resize domain technique then apply ABC for it.

Firstly we resize [ ]1, 2 +→ ¡ by the way: 1 11

x ax

→ − =−

.

Then take 2 2 2

, , , , ,1 1 1

a b cx y z a b c

a b c++ + += = = ∈

+ + +¡ . We need to prove this problem:

2 2 2 1 1 1( , , ) 10

1 1 1 2 2 2a b c a b c

f a b ca b c a b c

+ + + + + +� �� = + + + + ≤� �� + + + + + +� ��

.10

( 1)( 1)( 1)( 2)( 2)( 2)M N

a b c a b c⇔ ≤

+ + + + + +

where ( )2 ( 1)( 1) ( 1)( 2)( 1) ( 1)( 1)( 2)M a b c a b c a b c= + + + + + + + + + + +

and ( )1 ( 2)( 2) ( 2)( 1)( 2) ( 2)( 2)( 1)N a b c a b c a b c= + + + + + + + + + + +

Is it a big hurt when the inequality: 10( 1)( 1)( 1)( 2)( 2)( 2) 0MN a b c a b c− + + + + + + ≤ can not

solved by ABC theorem easily, because the coefficient of 222 cba will be negative (the

reader can easily mental calculate it). However the ABC Theorem now has more than it was

be, we can apply ABC Extra to solve this case. Fix abc and a b c+ + , and we will se that the

expression obtained is the first degree polynomial with variable .cabcab ++ Therefore we

can conclude that the inequality obtain its maximum in the case ( )( ) ( ) 0a b b c c a− − − = .

Then we only need to prove the inequality: ( ) ( ) ( )2 2 1 1, , 2 2 101 1 2 2

a b a bf a a ba b a b

+ + + += ⋅ + ⋅ + ≤+ + + +

( ) ( ) ( ) ( )2 2 1 12 2 0 3 2 3 2 01 1 2 2

a b a b ab b ab aa b a b

+ + + +⇔ ⋅ − ⋅ − ≥ ⇔ + + + + ≥+ + + +

So the problem has been proved completely.

Following is some apply problems:

Problem 1. Give [ ], , 1, 2a b c ∈ . Find the maximum of: 3 3 3

3a b c

Aabc

+ +=

Problem 2. [VN TST 2006]

Give [ ], , 1, 2x y z ∈ . Prove that: ( ) 1 1 1 6yx zx y z

x y z y z z x x y� �� + + + + ≥ + +� � � �+ + +� � � �


104

VI. THE ABC GENERALIZATION

So we have dealt with two weak points of ABC . The third weak points we can easily find out

is that ABC can only apply for three variables problems, so how about many variables

problems ??? . We will together solve this problem to come out with the ABC generalization.

Firstly, we should be familiar with some concept used in the ABC generalization.

1. ABC-able

a. Definition:

Consider a three variables problem ( )cbaf ,, .

We call ( )cbaf ,, an ABC-able expression if the ( ) 0,, ≥cbaf can be applied ABC theorem

to refer to two cases:

i) ( )( )( ) 0a b b c c a− − − = .

ii) 0abc = (This condition only occurs when apply ABC-able for variable abc )

b. An approach using ABC-able:

To prove that an expression ( ), ,f a b c is ABC-able, we will transform the expression

( ), ,f a b c to expression g with three variables , ,A a b c B ab bc ca C abc= + + = + + = . The

expression ( ), ,f a b c is an ABC-able expression if ),,( CBAg is a convex function for

variable BA, or C .

Example: Prove this expression is ABC-able

3 3 3( , , ) 3 ( ) ( ) ( )f a b c a b c abc ab a b bc b c ca c a= + + + − + − + − +

Solution

Take , ,A a b c B ab bc ca C abc= + + = + + = , we have:

3 3( , , ) ( , , ) 3 3 3 3 4 9f a b c g A B C A AB C C AB C A AB C= = − + + − + = − +

Consider ),,( CBAg only with variable C . We have: 9, 0g g′ ′′= = therefore g is a convex

function with variable C. Therefore ( ), ,f a b c is ABC-able.

2. The ABC Generalization

Consider a symmetric n variables function ( )1 2, ,..., nf a a a , where ƒ has minimum and 3n ≥ .

We will consider ( )1 2, , ..., nf a a a as a three variables function ( )321 ,, aaag and naaa ,...,, 54

are considered as constant.

If g is ABC-able then inequality ( ) 0,...,, 21 ≥naaaf can be refer to two cases :

)i m variables are equal and the rest n m− variables are equal.

)ii One variable is equal to 0 (This condition only occurs when apply ABC-able for variable abc )

The inequality is completely proved if it is correct in two cases given.


105

Proof

We will assume that ƒ has minimum as given and its minimum occurs when ( )nxxx ,...,, 21 (*)

If 1 2 ... 0nx x x = or ix can only obtain one in two fixed values then our theorem was proved.

If ix can obtain at least three different non-zero values, we assume that ( )321 ,, xxx are a set

that its elements are different in pairs and different from 0. We will fix nxxx ,...,, 54 as

constant and consider the function ( )nxxxxxf ,...,,,, 4321 as the function ( )321 ,, xxxg .

As given in theorem, g is ABC-able, so it get it minimum value only when

( )( ) ( )1 2 2 3 3 1 0x x x x x x− − − = or 1 2 3 0x x x = . This also means that there exists a set ( ), ,a b c

so that ( ) ( )1 2 3, , , ,g x x x g a b c> or ( ) ( )1 2 3 4 4, , , , ..., , , , , ...,n nf x x x x x f a b c x x≥ .

This is a contradiction with (*). In conclusion, the ABC generalization has been proved.

3. Problems with the ABC generalization

Problem 1 [Nguyen Anh Cuong]:

Prove that ( ) ( )4 4 4 4 2 2 2 23 3

14

a b c d a b c dabcd ab ac ad bc bd cd

+ + + + + +≥ ++ + + + +

, ∀ , , ,a b c d > 0 (1)

Solution

The inequality (1) ⇔ ( , , , )f a b c d = ( ) ( )4 4 4 43 a b c d ab ac ad bc bd cd+ + + + + + + +

( ) ( )2 2 2 24 12 0abcd ab ac ad bc bd cd abcd a b c d− + + + + + − + + + ≥ (2)

Notice that if we fix d then ƒ is a three variables symmetric polynomial, more than that ƒ is

ABC-able. Therefore according to ABC Theorem we only need to consider two cases (we

will consider the inequality in the form (1)):

i) 0a = : The inequality is trivial.

ii) ,a b x c d y= = = = :

Then (1) ⇔ ( ) ( ) ( ) ( ) ( )2 2 24 4 2 2

2 2 2 2 2 2 2 2

3 6 3 21

2 4 2 4

x y x y x y x y x y

x y x y xy x y x y xy

+ + + − −≥ + ⇔ ≥

+ + + +

Apply the inequality 2( ) 4x y xy+ ≥ ; 2 2 2x y xy+ ≥ we will prove the above inequality.

iii) ,a b c x d y= = = = :

Then (1) ⇔ ( ) ( ) ( ) ( ) ( )2 24 4 2 2 2 2

3 2 3 2

3 3 3 3 3 3 2 31

4 3 3 4 3 2

x y x y x y x xy y x y

x y x xy x y x xy

+ + − + + −≥ + ⇔ ≥

+ +

Apply the inequality 2 23 2 3x xy x+ ≥ and 2 23 2 4x xy y xy+ + ≥ we will prove for this case.

In conclusion, we have completed the proof.


106

Problem 2 [Nguyen Anh Cuong] Give 1 2, , ..., 0na a a ≥ satisfying 1 2 ... na a a n+ + + = .

Prove that: 2 2 2

1 2 1 2

1 1 1 2 1... 2 1...n n

n n n na a a a a a

−+ + + + ≥ + −+ + +

(1)

Solution

Take ( ) ( )1 2, , ..., 1nf a a a LHS= . Fix naaa ,...,, 54 we will obtain a symmetric three variables

expression which is also ABC-able. Therefore we only need to consider two cases:

i) For 1 0a = : The inequality is trivial.

ii) m variables are equal to x and (n – m) variables are equal to y, we have: ( )mx n m y n+ − = .

Prove that: 2 2

2 12 1

( )m n m n n

n nx y mx n m y

− −+ + ≥ + −+ −

Refer to: ( )

( )

2

2 2

2 1( )2 1

n mx n m ymx n m y m n m n nx yn n mx n m y

� �− + −+ − � �−� �+ + ≥ + −� �� + −� �

( ) ( ) ( ) ( )( )

( )2 2

22 22 2

2 1( ) 2 1 0

m n m x y n m n m x ymx n m y n xy x y

nxy nmx n n m y

− − − − −� �⇔ ≥ ⇔ + − − − − ≥� �+ −

The inequality is correct because: ( ) ( )2 2 2 2 1mx n m y m n m xy n xy+ − ≥ − ≥ − .

Therefore the problem was completely proved.

Problem 3 [Nguyen Anh Cuong] Give 1 2, , ..., 0na a a ≥ . Prove that:

( )2

1 1 21 23

1 2

.....

( ... )

i ji j n n n

nn

a aa a a

a a ann a a a

≤ < ≤−

+ + +≤ −

+ + +

�

Solution

We have met this problem in S.O.S technique; however this problem can not be solved completely using S.O.S. Here we will give a better solution, using ABC Generalization.

Letting ( )( ) 2

11 21 2 1 1 2

1 2

..., , ..., , ... 2

...

i ji j nn n

n n nn

a aa a a

f a a a a a a a nn a a a

≤ < ≤−

−+ + +

= − −+ + +

�

Here fixing 1 2 ... na a a+ + + and 1

i ji j n

a a≤ < ≤� is not a good idea because the degrees of 1 2 ... na a a

will be too big. For this reason we will fix 1 2 1... na a a −+ + + and 1 2 1... na a a − , clearly that ƒ will

become a quadratic trinomial with variables 1 2 1, ,... na a a − or a linear polynomial with variable

1i j

i j n

a a≤ < ≤� . Therefore ƒ is ABC-able with any three variables in the set { } 1

1

ni i

a−

=.


107

Here we prove that the expression is ABC -able with variable ab bc ca+ + so we have only one case to consider: m variables are equal, the rest n m− variables are equal.

Assume that ( )1 2, ,..., ( , ,..., , , ,... )n n n n n nna a a x x x y y y= (m variable x, n m− variable y).

Assume that x ≥ y (x ≤ y are consider similarly). We refer to the problem:

Give x, y ≥ 0. Prove that: ( )

( )( )

2

3 3

( ) n n n nm n m

n n

m n m x y mx n m yx y

nn mx n n m y−− − + −≤ −

+ −

Due to the homogeneous property, we can consider 1, 1y x= ≥ and we will need to prove that:

( ) ( ) ( ) ( )( )2 22 3 1 0n m n nf x n mx n m n x mx n m m n m x= + − − + − − − − ≥

We have referred to a one variable problem, but to prove it is still a big hurt.

Actually we must face a problem is that variable m is dynamic, non-continuous in [ ]0,n …

Through the first and second example, we have a notice that m may “prefer” 0, 1, 1,n n− .

A forsee feeling is that we may always push m to these values, then how easy our work will be …

And so great it is, this foresee is really correct. I will represent this result to the readers:

Result 1: In ¡ or +¡ , when two values 1 2,M a b c M ab bc ca= + + = + + are fixed then

3M abc= get it maximum value when ( ) ( ), , , ,a b c x x y= or its permutation then x y≤ , and if 3M

get it minimum value when ( ), , ( , , )a b c z z t= or its permutation then z t≥ .

Proof

We will consider the case when 3M get its maximum or minimum value when there are two

equal variables and refer to the problem:

yxM += 21 and xyxM 222 += . Find the maximum value of: yxM 2

3 = .

Replace xMy 21 −= , we will have: 023 212 =+− MxMx and

( ) ( )22 1 1 22 3 2

3 1 1

2 32 2

9

M M x M MM x M x x M x

− += − = − + = (notice that 2

2 13 0M M− ≤ )

Therefore, the maximum and minimum value of 3M depends only the maximum and

minimum value of x. Here more specify is that 3M will get its maximum and minimum value

when x get minimum and maximum values in order:

21 1 2 1

3

3 3

M M M Mx y

− −= ≤ ≤ and

21 1 2 1

3

3 3

M M M Mx y

+ −= ≥ ≥ .

With a same idea, we also hope for this result:


108

Result 2: In +¡ , when two values 1 3,M a b c M abc= + + = are fixed then the value

2M ab bc ca= + + get its maximum value when ( ) ( ), , , ,a b c x x y= or its permutation then

x y≤ , get its minimum value when ( ), , ( , , )a b c z z t= or its permutation then z t≥

Proof

In the proof for result 1, we have obtained this equality:

( )2 2 32

2 1 1 2 3 1 3 113 2

1 1

2 3 9 2 27

9 6 3 6

M M x M M M M x M MMM M

x M x M

− + + −= ⇔ = = +

+ +

Notice that in +¡ then 33 127 0M M− ≤ so 2M get its maximum and minimum values when x

get its minimum and maximum values in order. Also notice that x is a nonnegative solution of

the equation: 3 21 32 0x M x M− + = . Assume that this equation has three solutions: 0a b c≤ ≤ ≤ .

So 2M get its maximum and minimum values when x get values b and c in order.

We will prove that 1

3M

b c≤ ≤ to deduce that cyb ≤≤ .

Indeed, we have: 1 1

2 2M M

a b c a b c= + + � = − − . Replace this value a into the expression:

0ab bc ca+ + = we will obtain that: ( ) ( )( )

2 21 1 20

2 33M M b c bcb c b c bc

b c� � + ++ − − + = ⇔ =� � +

.

From here we can come out with a conclusion 1

3M

b c≤ ≤ and also obtain what we want to have.

Result 3: In +¡ , when two values 2 3,M ab bc ca M abc= + + = are fixed then the value

1M a b c= + + get its minimum value when ( ) ( ), , , ,a b c x x y= or its permutation then x y≥ .

Proof

We have 31 2

2M

M xx

= + where x is solution of the equation 32 32 0x M x M− + = . This

equation has two positive and one negative. Let call them 0a b c≤ ≤ ≤

We have: 0a b c a b c+ + = � = − − ; 3 3( ) 2 ( )abc b c bc M bc b c M= − + = − � + =

From here we can find out that: cMb ≤≤ 3 or cyb ≤≤ .

Now we will see ( )1M b or ( )

1M c are minimum value of 1M . We have:

( )1 1( )M b M c− ( ) ( ) ( ) ( )22 2 2 2

3 32 2 2 2 2 2

( )( ) 2 ( ) 22 0

M c b c b b c M b c b c bc b cb c b c b c

b c b c b c

− + − + − += − + = − = − ≥

So 1M get its minimum when ycx ≥= .


109

From these three result, we can upgrade out ABC Theorem as:

ABC Generalization-Extra

Consider a symmetric n variables function ( )1 2, ,..., nf a a a , where ƒ has minimum and 3n ≥ .

We will consider ( )1 2, , ..., nf a a a as a three variables function ( )321 ,, aaag and naaa ,...,, 54

are considered as constant.

If g is ABC-able with monotonic property then inequality ( ) 0,...,, 21 ≥naaaf can be referred

to two cases:

)i 1n − variables are equal.

)ii One variable is equal to 0 (This condition only occurs when apply ABC-able for variable abc )

The inequality is completely proved if it is correct in two cases given.

* ABC-able with monotonic property means g ′′ for one of three values , ,abc ab bc ca a b c+ + + + is 0.

Proof

According to the old ABC Generalization then we can refer to the problem when one variable is equal to 0, or m variable are equal to a and the rest mn − variable are equal to b.

In the first case, if there is one variable equal to 0 then there is nothing much to say.

In the second case, we need to prove that 1≤m . Indeed assume that 2≥m .

Notice that if g is a ABC -able function with monotonic property then the minimum value ƒ

only can be obtained when one of three values , ,abc ab bc ca a b c+ + + + get its maximum or

minimum values (do not need to consider both maximum and minimum). Let assume that ƒ get its minimum value when abc get its minimum (the rest cases are very similar).

WLOG, assume that ba ≥ and bxaxax === 321 ,, . Clearly when we fix all rest variables

nxxx ,...,, 54 then ( )baa ,, can not be the point for the function ),,( 321 xxxf get its minimum

value; because as we proved in the results this minimum can be obtained at the point ( )yxx ,,

where: yx ≤ .

This contradiction prove that m can not be greater than 1, and therefore there are 1−n variables are equal.

So we have completely the proof for the upgrade of ABC Generalization− .

We will return the problem 3 that we leaved before.

Notice that in this problem we have used ABC-able for the monotonic property of

cabcab ++ . Therefore, we only to consider one case when there are 1−n variables are equal,

and refer to the problem (which is the case when 1−n are equal):

( ) ( ) 222 3 1( 1) 1 ( 1) 1 ( 1) 1 0n n n nf x n n x n x n x n x−� � � �= − + − − + − − − ≥� � � �


110

( )[ ] ( ) ( ) 22 21 1 ( 1) 1 1 1 0n n n nn n x n x nx n x−� �⇔ − + − + − − − − ≥� � (*)

Notice two equality: ( )( )1 21 1 .. 1n n nx x x x− −− = − + + +

and ( ) ( ) ( ) ( )[ ]22 2 31 1 1 1 2 ... 2 1n n n nn x nx x n x n x x− − −− + − = − − + − + + + . Therefore:

(*) ⇔ ( ) ( ) ( )22 2 2 3 1 21 ( 1) 1 ( 1) 2 ... 2 1 ( 1) ... 1 0n n n n nx n n x n x n x x n x x− − − −� �� − − + − + − + + + − − + + + ≥� ��

Notice that: ( )[ ]( ) ( ) 2

211 1 1 11

nnn x x

n− + + ≥ +

−

( ) ( ) ( )22 3 12 3 2 2 21 1 11 ( 2) ... 2 1 ... 1 ... 1

1 2 2

n nn nn x n x x x x x

n n

− −− −� �− + − + + + + + + + ≥ + + + +� � − −

Therefore: ( )[ ] ( ) ( )[ ]2 2 31 1 1 2 ... 2 1n n nn n x n x n x x− −− + − + − + + +

( ) ( )( ) ( )2 22 2 3 1

2 2 2 21 .. 11 1 1 11 1 ...

1 2 2 1

n n nnx x x x

n n n

− −≥ + + + + +

+ + + + +− − −

( ) 22 1( 1)1 ...

1 11 ...2 1

n nn nx x

n

− −−≥ + + ++ + +

−

( ) ( ) ( )2 22 1 2 111 ... ( 1) 1 ...n n n nn n

x x n x xn

− − − −−≥ + + + = − + + +

� ( ) ( ) 22 2 3 1 2( 1) 1 ( 1) 2 ... 2 1 ( 1) ... 1 0n n n n nn n x n x n x x n x x− − − −� �� − + − + − + + + − − + + + ≥� ��

So the proof was completed.

Problem 4. Given dcba ,,, > 0 satisfy 1=abcd . Prove that

( ) ( ) ( ) ( ) ( )68 2 2 2 22 1 1 1 1a b c d a b c d+ + + + ≤ + + +

Solution

Let considering ( ) ( ) ( ) ( ) ( )68 2 2 2 2( , , ) 2 1 1 1 1f a b c a b c d a b c d= + + + + − + + +

( ) ( ) ( ) ( ) ( )2 28 2 2 2 2 62 1 2 2 1 ( )a b c ab bc ca ab bc ca abc a b c a b c d a b c d� �= + + + + + + − + + − + + − + − + + +� �

If we consider above function with the variable cabcab ++ then the inequality 0),,( ≤cbaf

is ABC -able. Thus we just need to consider the following problem:

Given x, y > 0 satisfy 3 1x y = . Prove that ( ) ( ) ( )3 68 2 22 1 1 3x y x y+ + ≤ +

The inequality ( )638 2

6 31 12 1 1 3x xx x

� � � �⇔ + + ≤ +� � � �


111

Problem 5. Give positive numbers dcba ,,, satisfying: 1=abcd . Prove that

2 2 2 2

1 1 1 12

1 1 1 1a b c d+ + + ≥

+ + + + (China MO)

Solution

Take 2

12 11

xd

= − >+

. We need to prove: 2 2 2

1 1 11 1 1

xa b c

+ + ≥+ + +

( )( ) ( )( ) ( )( ) ( )( )( )

( ) ( ) ( )

( ) ( )

2 2 2 2 2 2 2 2 2

2 2

2 2 2 2 2

1 1 1 1 1 1 1 1 1

1 2 4 2 ( )

2 ( ) 2 ( )

a b b c c a x a b c

a b c ab bc ca ab bc ca abc a b c

x x a b c x ab bc ca x ab bc ca xabc a b c xa b c

⇔ + + + + + + + + ≥ + + +

⇔ + + + + + + − + + − + +

≥ + + + + + + − + + − + + +

( ) ( )221 ( ) ( 2) 2(2 )( ) 2(1 ) ( ) 1 0x ab bc ca x a b c x ab bc ca x abc a b c x⇔ − + + + − + + + − + + + − + + + − ≤

The above inequality is ABC able− with monotonic property when we consider the variable cabcab ++ . We only need to consider the case:

Give positive numbers 3, : 1x y x y = . Prove that: 2 2

3 1 21 1x y

+ ≥+ +

2

6

3 12

11 1xx

⇔ + ≥+ +

6

2 6

3 3 10

2 21 1x

x x⇔ − + − ≥

+ +

( ) ( )( )

( ) ( )( )

5 4 3 2

2 6

3 1 1 1 10

2 1 2 1

x x x x x x x x

x x

− + − + + + + +⇔ + ≥+ +

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 5 4 3 2 3

2 2 3 2

1 1 1 3 1 1 0

1 1 1 2 4 2 0

x x x x x x x x x

x x x x x x x

� �⇔ − + + + + + + − + + ≥� �

⇔ − + − + + + + ≥

The equality occurs ⇔ 1a b c d= = = =

Problem 6 [Vasile MS 2005] Give n positive numbers satisfying: 1...21 =naaa .

Prove that: 1 2 1 2

41 1 1... 2...n n

n na a a n a a a

+ + + + ≥ ++ + + +

(1)

Solution

(1) ⇔ 1 2 2 3 3 1

1 2 3 4 5 1 2 3 4

1 1 1 4... 2

...n n

a a a a a a nn

a a a a a a n a a a a a

+ ++ + + + + ≥ +

+ + + + + +

Consider the variable 133221 aaaaaa ++ , clearly that it is ABC able− with monotonous

property, also we use the minimum value of 133221 aaaaaa ++ so the equality variables

should be greater than the rest one. Therefore we only need to consider the inequality:


112

Give positive yx, satisfying 11 =− yx n and 1≥x , then:

1 41 2( 1)

n n nx y n n x y− + + ≥ +

+ − + 1

1

1 4 21( 1)

n

n

n nx nx n n x

x

−

−

−⇔ + + ≥ ++ − +

( )[ ]1

1

1 2 1 1( 1) 1

n n n

n n

x n nx n x nxx nx n x

−

−

+ − − − − +⇔ ≥+ − +

( ) ( )22 2 3 2 3

1

( 1) 2 ... 1 2 1 1 ( 2) ... 1

( 1) 1

n n n n

n n

x x x n x n x n x

x nx n x

− − − −

−

� � � �− + + + − − − + − + +� � � �⇔ ≥+ − +

We need to prove that: 2 3 2 3

1

2 ... 1 ( 1) ( 2) ... 1( 1) 1

n n n n

n n

x x n n x n xx nx n x

− − − −

−+ + + − − + − + +≥

+ − +

( )1

2 3 2 3( 1) 12 ... 1 ( 1) ( 2) ... 1

2

n nn n n nnx n x

x x n n x n xx

−− − − −� �+ − +⇔ + + + − ≥ − + − + +� �

Notice that because 1x ≥ then 1 ( 1) 1 ( 1) 1

12 2

n nnx n x nx n xn

x x

− + − + + − +≥ ≥ − . Hence:

LHS ≥ ( ) ( )2 3 2 31 2 ... 1 ( 1) ( 2) ... 1n n n nn x x n n x n x− − − −− + + + − ≥ − + − + +

So we have the needed conclusion.

Problem 7. Given nxxx ,...,, 21 > 0 satisfying: 1 2

1 1 1...n

nx x x

+ + + = . Prove that:

( )1 2 1 1 2... e ... 1n n nx x x n x x x−+ + + − ≤ − where ( ) 1

11e 1 e

1

n

n n

−

− = + <−

(Gabriel Dospinescu and Calin PoPa, MS, 1004)

Solution

Notice that we can rewrite the problem as:

Give nxxx ,...,, 21 > 0 satisfying: 1 2 2 3 3 1 1 2 34

1 1..

n

x x x x x x n x x xx x

� �+ + = − − −� �

.

Prove that: ( )1 2 3 1 1 2... e ... 1n n nx x x x n x x x−+ + + + − ≤ − where ( ) 1

11e 1 e

1

n

n n

−

− = + <−

Hence if consider the inequality with variable 321 xxx ++ , the inequality is ABC able− .

Therefore our work is only this problem: Give positive numbers nba

nba =+− 11

:, , then :

( ) ( )11( , ) 1 e 1n

ng a b n a b n a b−−= − + − ≤ − (*) where ( ) 1

11e 1 e

1

n

n n

−

− = + <−


113

Replace 1

abna n

=− +

into (*) , we need to prove that:

( ) 11( ) ( 1) 1 11 11

n na af a n a nn na nna n

− � �= − + − ≤ + −� �− − + − + where

1na

n−≥

( )

( ) ( ) ( )

2 2 2 1

2 1

2 2 3 0

( 1) (2 2 )11

( 1)( 1)11 2 ... ( 1) ... ( 1)

n

n

n

n n n k

n n a n n a n n nna na n

n n a nna a a k a n a

−

−

− − −

− − − + −⇔ ≤

−− + −

− −⇔ ≤

−− + + + − + + −

( ) ( ) 1

2 3 0

112 ... ( 1)

n

n n

n n nna a n a

−

− −−⇔ ≤

−+ + + −. Notice that:

1na

n−≥ .

Therefore: LHS ( )

( ) ( ) ( )( ) ( ) 1

2 3

1 111 12 ... 1

n

n n

n n n nfn nn n n

n n

−

− −

− −≤ = =−− −+ + + −

(**)

At (**), notice that ( )1nfn− , which is also the value of ( , )g a b when

1,

na b

n−→ → ∞ , or

exactly is: ( ) 1

1

nn

n

−

− (**).

4. Proposed Problem

Problem 4.1. [Nguyen Anh Cuong] Give 0,,, >dcba . Prove that:

( ) ( )4 4 4 41 1 1 112

a b c da b c d

a b c dabcd+ + + + ≥ + + + + + +

Problem 4.2. Give positive numbers naaa ,...,, 21 satisfying: �≤<≤

=nji

ji aa1

1 .

Prove that: ( )22 2

1 1 22 2... ... min 2,

11

n

n na a ka a a kn nn

�� + + + ≥ + � �−− � � � �

Problem 4.3. Give nxxx ,..,, 21 satisfying 1... 222

21 =+++ nxxx . Prove that:

( ) ( )3 3 31 2 1 2 3

6... 1

2 1nx x x x x x

n n+ + + + ≤

− +�

Problem 4.4. [Vasile Cirtoaje, MS, 2004] Given nxxx ,...,, 21 > 0. Prove that:

( )1 2 1 21 2 1 2

1 1 1 1... ... ... 2...n n

n n

x x x n n x x xx x x x x x

� �+ + + − + + + − + + ≥� �

Problem 4.5. [Vasile Cirtoaje, MS, 2006]

Give 0,...,, 21 ≥nxxx satisfying: 1 2 ... nx x x n+ + + = . Prove that:

( ) ( )12 2 2

11 2 1 2... ...nn nx x x x x x n− + + + ≤


114

VII. ABC CYCLIC

The fourth weak point and also quite a hot problem in these days is cyclic problem. For now, there are no method which is really effectible for cyclic inequality.

1. Cyclic to symmetric with variable transformation

For some cyclic inequality, we only need a variable transformation step to refer to a symmetric problem. Notice that the variable transformation also should have the cyclic property, because if we still keep the symmetric property then we can never refer a cyclic to symmetric. Let consider some example:

Problem 1 [Bodan-Mathlinks]

Give real numbers 0,, >zyx . Prove that: 1

12 2 2 2

yx zx y y z z x

≥ + + >+ + +

(1)

Solution

(1) ⇔ 1 1 1 1

122 22

y z xy zx

≥ + + >+ ++

. Take , ,y xza b cx y z

= = = � 1abc = .

Then the inequality becomes: 1 1 1 1

12 2 2 2a b c

≥ + + >+ + +

The readers can also see immediately that when we transform the above inequality to polynomial form then it is still 3-degrees with cba ,, , therefore linear polynomial with

variable cabcab ++ . Therefore fix cba ++ and abc , and we need to consider the only case

when ( ) ( ) ( ) 0a b b c c a− − − = :

The problem becomes: Give , 0a b > and 2 1a b = . Prove that: 2 1 1

12 2 2a b

≥ + >+ +

Replace 2

1b

a= and refer to the problem:

2

2

2 11

2 22 1a

a a≥ + >

+ +

The left inequality is equivalent to: ( )21 0a a − ≥

The right inequality is equivalent to: ( )22 74 1 0 2 1 02 2a aa a− + > ⇔ − + >

So we have completed the proof. The equality occurs ⇔ 0x y z= = >

Problem 2. Give 0,, >cba . Prove that: 2 2 2

65

a b c abcb c a a b b c c a

+ + + ≥+ +

(1)

Solution

Take ; ;a b c

x y zb c a

= = = � 1=xyz . Then (1) ⇔ 6

5x y zxy yz zx

+ + + ≥+ +

(2)

This is a simple problem with ABC extra− .


115

We refer problem to the only case ( )( )( ) 0x y y z z x− − − = , assume that zy =

� 22

11xy xy

= � = . Then (2) ⇔ 2

62 5

2x y

y xy+ + ≥

+

2 3

612 5 0

2y

yy y

⇔ + + − ≥+

( ) ( ) ( )2 2

2 3

2 3 1 2 10

( 2)

y y y y y

y y

� �+ − + + −� �⇔ ≥+

(is always true). The equality occurs ⇔ 0a b c= = >

* In the above problems, we have introduced some technique to transform a cyclic to

symmetric problems using variable transformation, and then use ABC to handle the rest. Now

we will introduce another solution for cyclic case. With this idea, we can handle almost cyclic

polynomial of which degrees is less than or equal to 5 in ¡ .

We will consider some example:

Problem 4. Prove that: 4 4 4 3 3 33( ) 4( ) 0a b c a b b c c a+ + + + + ≥ , , ,a b c∀

Solution

Here we will deal with a stronger problem: 4 4 4 3 3 3197( ) 280( ) 0a b c a b b c c a+ + + + + ≥ (1)

Take 2 7 ; 2 7 ; 2 7a x y b y z c z x= + = + = + . The inequality (1) is equivalent to:

( )4 4 4 3 3 3 3 3 3

2 2 2 2 2 2

( , , ) 222743( ) 240296( )

92904 246960 ( ) 0

f x y z x y z x y y x y z z y z x zx

x y y z z x xyz x y z

= + + + + + + + + +

+ + + + + + ≥

),,( zyxf is a symmetric polynomial, more than that it also a homogenous polynomial of

which degree is four. We only need to consider one case:

( ) 4 3 2,1,1 0 222743 480592 432768 974512 1018982 0f x x x x x≥ ⇔ + + + + ≥

which is no longer difficult and will be remain for the readers.

This solution using ABC which is quite clear and nice, however the mystery inside it is that we

have transform variable to refer to a symmetric problem 2 7 ; 2 7 ; 2 7a x y b y z c z x= + = + = + .

It is not quite trivial to go to this transformation; we need some small testing and convention

to find out. Our target is that we will make sure that all coefficients for cyclic expression will

be equal after the transformation. More specify here we take: kxzckzybkyxa +=+=+= ,,

and find k so that two coefficient of two cyclic expression 3 3 3 3 3 3,x y y z z x xy yz zx+ + + +

are equal. For the general problem: ( )4 4 4 3 3 3 0x y z m x y y z z x+ + + + + ≥ , we need to find k

such that: 3 2( 4) (4 2 ) 0mk m k m k m− + + − + = .

For some case to find a good coefficient for the cyclic pair expression which as 4 4 4x y y z z x+ + , 4 4 4xy yz zx+ + and 3 2 3 2 3 2 2 3 2 3 2 3,x y y z z x x y y z z x+ + + + we also need to take:

, ,a x my nz b y mz nx c z mx ny= + + = + + = + +


116

So for now the readers might also guess that why this idea is only effectible for polynomial of

which degrees are not greater than 5, from 6-degrees or above we have more than two cyclic

expressions , for example with 6-degrees is 5 5 5 5 5 5,x y y z z x xy yz zx+ + + + ,

4 2 4 2 4 2 2 4 2 4 2 4,x y y z z x x y y z z x+ + + + and 2 2 2 2 2 2( ), ( )xyz x y y z z x xyz xy yz zx+ + + +

so we may not find a good system to find out nm, .

We will consider some another example to understand better this technique:

Problem 5 [MnF]. Give positive numbers cba ,, . Prove that:

( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 2 2 2 2 2 2 213 10 5 9 13 10 5 9 13 10 5 9 0a ab b c a b b bc c a b c c ca a b c a− − + − + − − + − + − − + − ≥

Solution

The inequality is equivalent to: ( )( ) ( )( )[ ]22 4 2 4a b a b b c b c− − − − − +

[ ] ( ) 22( 2 )( 4 ) ( 2 )( 4 ) 2 ( 4 ) ( 2 )( 4 ) 0b c b c c a c a c a c a a b a b� �+ − − − − − + − − − − − ≥� �

This problem is a challenge problem more than a nice problem. Its author has tried for this

inequality have two separated equality point a b c= = and 2 4a b c= = . However if we have

known about ABC-cyclic then this is a good problem to apply:

Take 2 , 2 , 2a x y b y z c z x= − = − = − . The inequality is equivalent to:

( ) ( ) ( )( )

4 4 4 2 2 2 2 2 2

3 3 3 3 3 3

, , 4 21

10 5 ( ) 0

f x y z x y z x y y z z x

x y xy y z yz z x zx xyz x y z

= + + + + +

− + + + + + − + + ≥

This is a fourth degrees polynomial and also a homogeneous one in ϒ, we only need to

consider one case: ( ) ( ) ( )2 24 3 2,1,1 4 20 37 30 9 1 2 3 0f x x x x x x x= − + − + = − − ≥ .

This problem is more special than the problem 3 in the point that it has two separate equality

point . Sometime this case make it easer to find out the k .

For example we assume that: , ,a x ky b y kz c z kx= + = + = + . When the maximum or minimum

occurs we should have two variable in , ,x y z equal; let assume that 1y z= = and then:

, 1, 1a x k b k c kx= + = + = + . From here, combine with this system:

2a b= and 4a c= we can easily find out that 31 ,2 2

k x−= = or 1,1 =−= xk .

From here, try in the inequality and we come out with the transformation given in the solution.


117

In above examples, all the problem has the domain is ¡ , however almost the problem will be

in +¡ , and more specify is that if its degree is odd then we must consider in +¡ . So should

we surrender in these cases? Actually it is not, let consider this example:

Problem 6 [Nguyen Anh Cuong]: Given 0,, ≥cba . Prove that

( ) ( ) ( )3 3 3 2 2 2 2 2 223 17 37 9a b c a b b c c a ab bc ca abc+ + + + + ≥ + + +

Solution

We again use the idea taking , ,a x ky b y kz c z kx= + = + = + . For this problem, the equality occurs

in two separate points, and one of these can be specified in the case 1, 1, 2a b c= − = = and its

permutation. In this case in three variables zyx ,, , there are two variable should be equal; let

assume that zy = .

Then we will have the system: 1; ( 1) 1; 2x ky k y kx y+ = − + = + = . From here we can find out k

through this equation: 22 3 1 0k k+ + = , solve it to obtain 1k = − or 12

k = − .

Choose 12

k = − , and take: 2 , 2 , 2a x y b y z c z x= − = − = − and the problem will become:

3 3 3 3 ( ) ( ) ( )x y z xyz xy x y yz y z zx z x+ + + ≥ + + + + + which is Shur inequality, can be solver by

many ways or ABC. However the above inequality only valid for 0,, ≥zyx . Fortunately,

4 2 4 2 2 4, ,7 7 7

a b c a b c a b cx y z+ + + + + += = = are non-negative so the proof should be end here.

The variable transformation idea is really a good point to remember, however it is not a

perfect solution anyway. Following we will introduce a more advance idea which should be

perfect in some meanings.

2. The connection between cyclic and transformation

The readers may sometimes ask yourself if the symmetric is only “the sun” of cyclic

inequality, because if the cyclic inequality is correct than we can refer to the symmetric one

easily. This is correct in some meaning, for example this problem:

2 2 2

2 2 23

a b c a b c b c ab c a a b cb c a

+ + + ≥ + + + + + is correct, then the following symmetric

problem is also correct: 2 2 2 2 2 2

2 2 2 2 2 22 3 2

a b c b c a a b c b c ab c a a b cb c a a b c

� �+ + + + + + ≥ + + + + +� �

However the reverse question, if the symmetric one is correct then how about the cyclic one. A

quick thinking come out with the answer is NO. For example:


118

( ) ( ) ( )3 3 3 3a b c abc ab a b bc b c ca c a+ + + ≥ + + + + + is correct but its cyclic inequality is not

correct: ( )3 3 3 2 2 23 2a b c abc a b b c c a+ + + ≥ + + is not valid for ( ) ( ), , 0,3, 2a b c = .

However I claim that every cyclic inequality has its equal symmetric. And it is enough to

solve the equal symmetric form to prove that the cyclic one is also correct. I will represent

this technique and its application as following:

SYMMETRIC TRANSFORMATION

Symmetric transformation technique for cyclic:

Give a function with variable cba ,, satisfying: ( ) ),,(),,(,, bacfacbfcbaf == .

If we take ( ) ( ) ( ), , , , , ,g a b c f a b c f a c b= ; ( ) ( ) ( ), , , , , ,h a b c f a b c f a c b= + then hg, are

symmetric inequality. Also the inequality ( ) 0,, ≥cbaf are equivalent to: ( )

��

≥≥

0),,(0,,

cbah

cbag

The idea seem simple but let see its power when we combine it with ABC :

Problem 6. Give 0,, ≥cba satisfying: 3=++ cba . Prove that: 2 2 2 4P a b b c c a= + + ≤

Solution

Take: 4),,( 222 −++= accbbacbaf

And: ( ) ( ) ( ) 1612)1839(9,,,,,, 32 +−+−+== BBCBCbcafcbafcbag

( ) 8)()(),,(),,(),,( −+++++=+= accacbbcbaabbcafcbafcbah

Where: .; abcCcabcabB =++= What we need to do is prove:

i) ( ), , 0g a b c ≥ ii) ( ), , 0h a b c ≤

We will prove the second one first, since this is quite simple. The readers can also see that

0),,( ≤cbah very quickly using ABC .

However here we suggest a shorter solution using Schur:

[ ] [ ]

( ) [ ]

3 3 3

3 3 3

3

3 ( ) ( ) ( )

6 3 ( ) ( ) ( ) 3 4 ( ) ( ) ( )

4 ( ) ( ) ( )

a b c abc ab a b bc b c ca c a

a b c abc ab a b bc b c ca c a abc ab a b bc b c ca c a

a b c ab a b bc b c ca c a

+ + + ≥ + + + + +

⇔ + + + + + + + + + ≥ + + + + + +

� + + ≥ + + + + +

( )3 27( ) ( ) ( ) 8

4 4a b c

ab a b bc b c ca c a+ +⇔ + + + + + ≤ = < � ( ), , 0h a b c ≤


119

Now the heavier part is the second part, in this part I have represented it in the form

,ab bc ca abc+ + to apply ABC for it. Here we will try to prove that g is a monotonic function

with variable C by using derivative and prove that the derivative is positive or negative.

Let consider it:

Take: ( ) 22 3 2( ) 9 (39 18 ) 12 16 9 (39 18 ) 2 ( 4)C C B C B B C B C B Bω = + − + − + = + − + − +

( ) 18 39 18C C B′ω = + −

First notice that if 3918

B ≤ then everything is good for now, because 39 18 0B− ≥ so ( )C 0ω ≥ .

Hence just consider the case when 3918

B > .

Notice that for every fixed value of B , ( ) 18 39 18 0C C B′ω = + − = has solution 18 3918

BC −= ,

or has no solution.

For the first case, we have: ( ) ( ) 3 2 10518 39C 9 2718 4

B B B B−ω ≥ ω = − + − .

And now it is not so difficult to prove that 3 2 1059 27 0

4B B B− + − ≥ for 39

18B >

In the case ( ) 18 39 18 0C C B′ω = + − = has no solution, means ( )w C is monotonous, we refer

to the original problem with the case ( ) ( ) ( ) 0a b b c c a− − − = or 0abc = .

i) Given 0, ≥ba satisfying: 3=+ ba , prove that: 42 ≤ba

This can be done quickly using AM – GM inequality: ( )32 4 4 4

32 2a a a ba b b += ⋅ ⋅ ⋅ ≤ =

ii) Given 0, ≥ba satisfying: 32 =+ ba , prove that: 4322 ≤++ aabba .

Replace ab 23 −= , the inequality ⇔ 04993 23 ≤−+− aaa (is true for 302

a≤ ≤ )

We have completely finished our proof. This proof is not the best solution for this problem,

but it should be an effectible solution for these kinds of problems.

A question come out immediately is that is accbba 222 ++ get it maximum and minimum

value when there are two equal variable or one variable is equal 0 when we fix

cabcabcba ++++ , ? This is a very possible foresee, because for the above problem the

equality occurs when there is a zero variable. Let consider this problem:


120

Problem 7 [Nguyen Anh Cuong]. Give positive numbers cba ,, satisfying:

)(2222 cabcabcba ++=++ . Find the maximum value of: 2 2 2

3

8( )( )

a b b c c aP

a b c+ +=+ +

.

Solution

The first remark is that this is a homogeneous cyclic inequality. We will use these properties

step by step to come out with final solution:

Firstly, due to the homogeneous property, we can take 1=++ cabcab .

The problem becomes: Give positive numbers cba ,, satisfying: 1,2 =++=++ cabcabcba .

Find the maximum value of 222 cabcbaP ++= .

Still the same idea to the first problem, we will take 222 cabcabQ ++= . However here the

problem ask us to find the maximum and minimum value ourselves so it is hard to define g

and h . However this can not make us surrender, we still go on with this idea.

Take abct = , as what we proved for the first part of this article, we have 40,27

t � �∈� ��

.

We have: ( ) ( )2 2 2 2 2 2 3 2 3P Q a b b c c a ab bc ca a b c ab bc ca abc t+ = + + + + + = + + + + − = −

( ) ( ) ( ) ( )2 3 3 29 6 ( ) 9 4 1PQ abc a b c a b c ab bc ca c ab bc ca t t� �= + + + − + + + + + + + = − +� �

QP, can obtain one of these two values 2 22 3 27 4 2 3 27 4

;2 2

t t t t t t− − − + − + − +.

To find out the maximum value, we should consider the case:

22 3 27 4( )2

t t tP f t − + − += = for 40,27

t � �∈� ��

. We have ( ) 0f t′ = have the unique root 127

t = .

Therefore ( ) ( ){ }max1 4max (0), ,27 27

f f f f= . From here we have max109

f =

when 127

abc = , when 0.04020492...; 0.78243211...; 1.17736297...a b c= = =

Hence our speculation was wrong. How pity is it ! We can not find out a beautiful thing such

as the symmetric world. Indeed the equality for cyclic inequality is always a troublesome but

also very amazing and joyful. You can also think more about this article to find out that why

almost cyclic inequalities get its maximum or minimum values when three variables are

completely different. We will stop the research for now; following is some example and

application problems using the symmetric transformation technique.


121

Problem 8 [Nguyen Anh Cuong] Give nonnegative cba ,, satisfying 4a b c+ + = .

Prove that: 2 2 2 2 2 23( ) 36 5( )a b b c c a ab bc ca+ + + ≥ + +

Solution

Take ( ) ( ) ( )2 2 2 2 2 2, , 3 5 36f a b c a b b c c a ab bc ca= + + − + + +

And: ( ) ( ) ( )( , , ) , , , , 72 2 ( ) ( ) ( )h a b c f a b c f a c b ab a b bc b c ca c a= + = − + + + + +

( ) ( ) ( ) ( )2 3 2, , , , . , , 441 4312 1176 64 240 288 1296g a b c f a b c f a c b C B C B B B= = + − + − − +

We will again use the same technique as the problem 7 to solve this problem.

Firstly prove that ( ), , 0h a b c ≥ . As the first problem,

( )3

( ) ( ) ( ) 16 364

a b cab a b bc b c ca c a + ++ + + + + ≤ = < hence ( ), , 0.h a b c ≥

Secondly is proving ( ), , 0g a b c ≥ .

Rewrite ( ), ,g a b c as ( ) 2 2441 (4312 1176 ) ( 3) (64 144)C C B C B Bω = + − + − +

So if 4312 111176 3

B ≤ = then we will have what we need immediately.

Consider 113

B > , we have: ( ) 882 4312 1176C C B′ω = + − , so using the same idea with the

problem 7 we need to prove three things:

i) ( ) 3 2 16384 8320012 44 64 1024 09 3 9

B B B B−ω = − + − ≥ when 113

B ≥ :

this is quite easy and we will skip it (this is the case where there is no equality)

ii) Give , 0a b ≥ satisfying: 2 4a b+ = . Prove that:

( ) ( )2 2 3 2 2 3 3 2 23 36 5 18a b b a a a b b a a a a b b a+ + + ≥ + + ⇔ ≥ + +

Replace 4 2b a= − and we have: 3 23 12 16 18 0a a a− + − ≤ where: 0 2a≤ ≤ , again remain for

the readers (this is the case where there is no equality)

iii) Give , 0a b ≥ satisfying: 4a b+ = . Prove that: 2 23 36 5a b ab+ ≥ .

Replace 4a b= − and we have: 3 2 28 44 48 36 0 ( 3) (8 4) 0b b b b b− + + ≥ ⇔ − + ≥

In conclusion, the inequality was proved completely.

The equality occurs when ( ), , (0,3,1)a b c = and its cyclic permutation.


122

Problem 9. [Vasile Cirtoaje]

Give real numbers: cba ,, . Prove that: ( ) ( )22 2 2 3 3 33a b c a b bc c a+ + ≥ + +

Solution

A very famous problem of Vasile Cirtoaje, a very amazing result in cyclic inequality world ,

a very strong result with the equality occurs at three separate points. And let see how the

power of symmetric transformation versus this powerful problem.

Firstly, because of its homogeneous property we need to have this condition 1a b c+ + = .

Take ( ) ( ) ( )22 2 2 3 3 3, , 3f a b c a b c a b b c c a= + + − + + and

( ) ( ) ( ) ( ) ( ) ( ) ( )22 2 2 2 2 2 2 2 2, , , , , , 2 3g a b c f a b c f a c b a b c ab a b bc b c ca c a� �= + = + + − + + + + +� �

( ) 2 2 4 3 2, , ( , , ) ( , , ) 63 (21 57 12) 49 68 42 11 1h a b c f a b c f a c b C B B C B B B B= = + − + + − + − +

Proving ( , , ) 0g a b c ≥ is in the border of ABC and will be remained for the interested readers.

Here we will give the proof for ( , , ) 0h a b c ≥ .

Take ( ) ( )2 2 4 3 263 21 57 12 49 68 42 11 1c c b b c b b b bω = + − + + − + − +

We have: ( ) 2126 21 57 12c c b b′ω = + − + . We will consider three cases:

i) ( )22 4 3 2 221 57 12 3969 4914 2277 468 36 63 39 6

0126 84 84

b b b b b b b b� �− + − − + − + − +ω = = ≥� �

ii) Two equal variables: Give the real numbers yx, . Prove that: ( ) 22 2 4 3 32 3( )x y x x y y x+ ≥ + +

Give 1x = and we refer to the problem: ( ) ( )24 3 2 23 4 3 1 0 1 1 0y y y y y y y− + − + ≥ ⇔ − − + ≥

iii) One variable is equal to 0: Give the real numbers yx, . Prove that: ( ) yxyx 3222 3≥+

This can be done quickly using AM – GM inequality: ( )22 2 2 2 3 3161 1 1 3

3 3 3 27x x x y x y x y+ + + ≥ >

Therefore the problem has been proved completely.

A remark about the power of this technique. Why this technique is powerful? The answer is

that because the equivalent through every transformation. The only trouble point is that with

which value of b then ( )c′ω will has solution or not. Actually this also can be solved by

ABC, because the derivative expression is also a linear polynomial with abc . This remark is

to see that, in some meanings, this technique is a possible approach to solve almost cyclic

polynomial inequalities no matter what.


123

Proposed Problems

Problem 5.1 [V�Q-MnF] Give 0,, >cba . Find the maximum value of k such that:

3( )( )( ) 8

a b c kabc kb c a a b b c c a

+ + + ≥ ++ + +

Problem 5.2. Find the maximum value of k such that:

( )2 2 2

2 2 23a b c a b ck k

b c ab c a+ + + ≥ + + valid for .0,, >∀ cba

Problem 5.3 Give positive cba ,, and a positive real number p. Prove that :

( )

2 2 2

2

3min 1,

1

a b ca pb b pc c pa p

� �� + + ≥� � � � � � � �+ + + +

Problem 5.4 [Nguyen Anh Cuong]

Give nonnegative real numbers cba ,, such that: 5=++ cba . Prove that:

( ) ( )3 3 3 3 3 316 640 11a b b c c a ab bc ca+ + + ≥ + +

Problem 5.5 [Vasile Cirtoaje]

Prove that: ( ) ( )3 2 2 24 27a b c ab bc ca abc+ + ≥ + + + , , , 0a b c∀ ≥

Problem 5.6. Give cba ,, > 0 satisfying: 1=abc .

Prove that: abcaccbba ≥++ 222


i) Give nonnegative numbers cba ,, satisfying: 3=++ cba .

Prove that: ( )( )2 2 2 9ab bc ca a b b c c a+ + + + ≤

ii) Give nonnegative cba ,, satisfying: 3=++ cabcab .

Prove that: ( )( )2 2 2 9a b c a b b c c a+ + + + ≥


Give nonnegative real numbers zyx ,, . Prove that:

i) ( ) ( ) ( )5 5 5 3 2 3 2 3 2 2 3 2 3 2 32 9 11x y z x y y z z x x y y z z x+ + + + + ≥ + +

ii) ( ) ( )5 5 5 4 4 4 4 4 42 3x y z x y y z z z xy yz zx+ + + + + ≥ + +


124

3. Cyclic to Symmetric using auxiliary inequality

In two previous parts, we have been familiar with cyclic inequalities and we solve it directly

in many ways. That is truly the strongest problems. However in these days people do not

always like it. We can use auxiliary inequalities to transform a cyclic inequality to a

symmetric inequality to apply ABC for it.

To make this idea clearer, let consider this example:

Problem 11 [Nguyen Anh Cuong] Give positive numbers cba ,, satisfying 3=++ cba .

Prove that: ( )( )2 2 2

2 2 2 93

a b ca b b c c a ab bc ca + ++ + + + ≤

Solution

Firstly let prove the auxiliary inequality:

Give positive numbers cba ,, satisfying: 3=++ cba . Then: 2 2 2 9a b b c c aab bc ca

+ + ≤+ +

.

This is actually the problem given in the symmetric transformation part, the poof is remained

for the readers. Here we will apply this inequality to transform the first cyclic problem to a

symmetric problem. So ( )( )2 2 2 9 ( ) 9a b b a c a ab bc ca ab bc caab bc ca

+ + + + ≤ + + =+ +

Our work is prove that: 2 2 2

9 93

a b c+ + ≥ .

This one is trivial by using: ( ) ( ) 22 2 23 a b c ab bc ca+ + ≥ + + .

And we can end the proof here.

Problem 12 [Nguyen Anh Cuong] Give positive problems , ,a b c satisfying: 3a b c+ + = .

Prove that: 2 2 2

23

a b c a b cb c a

+ ++ + ≥ +

Solution

This time we still use the same ideas, we need to have a good lower boundary for 2 2 2ab bc ca+ + .

Also in the proposed problems in symmetric transformation part we have this result:

Give nonnegative cba ,, satisfying: 3=++ cabcab . Then ( ) 2 2 2( ) 9a b c ab bc ca+ + + + ≥ .

Or rewrite as a homogeneous form is that: ( )( ) ( )22 2 2a b c ab bc ca ab bc ca+ + + + ≥ + + .

So with the condition 3=++ cba we have this inequality: ( ) 2

2 2 2

3

ab bc caab bc ca

+ ++ + ≥

We only need to prove that: 2 2 2 2( )

23 3

ab bc ca a b cabc

+ + + +≥ +


125

Consider the problem with variable abc , clearly this inequality is ABC able− .

We only need to consider two cases:

Case 1: 0=c , the inequality is trivially correct.

Case 2: ycxba === , , the inequality is equivalent to:

( ) ( )2 22 2 2 2 2

2

2 22 22 2

3 3 33

x xy x yx y x yyx y

+ ++ +≥ + ⇔ ≥ + where yx, are positive real

numbers satisfying 2 3.x y+ = Replace ( )3 2 , 0,1.5y x x= − ∈ we have:

( ) ( )2 226 3 2 3 22

3(3 2 ) 3x x x

x− + −≥ +−

( ) 2

23 23 2 4 3 1

3 2x

x xx

−⇔ − ≥ − + −−

( ) ( )( ) ( )22

2 2

2

3( 1) 2 13 2 1 1 4 3 1 0

3 2 2 4 3 1

x xx x x

x x x

− −⇔ ≥ ⇔ − + + − − ≥− − + +

The proof can end here.

Remarks: Through these two problems, the readers may also understand what that idea for

this method is. We will try to find good upper and lower boundary for cyclic expression using

(can use symmetric transformation) to transfer a cyclic problem to a weaker but nicer to

handle symmetric problem.

Let review two evaluation we have used in above two problems:

( ) ( )2 52 2 2 (*)

27( )ab bc ca a b c

a b b c c aa b c ab bc ca+ + + +≤ + + ≤

+ + + +

And another evaluation, which is exercise 5.02 and 5.03 in the symmetric transformation part.

( ) ( )33 2 2 2 2 2 2 427

a b c a b c a b b c c a a b c abc+ + ≤ + + ≤ + + − (**)

Also notice that we refer to the symmetric problem simply because we want to use ABC to

refer to the case where one variable is 0 or two variables are equal.. Therefore let consider (*)

and (**) in these two cases:

� When 0, 1c b= = , ( )52

2 1(*)

1 27a a

aa a

+⇔ ≤ ≤+

and ( ) ( )32 4 1

** 027

aa

+⇔ ≤ ≤

Notice that ( ) ( )3 52

2 4 1 10

1 27 27a a a

aa a

+ +≤ ≤ ≤ ≤+

, therefore in the case our inequality very near

to 0 when our variable reach 0 this evaluation should be:

( ) ( )2

32 2 2 427

ab bc ca a b b c c a a b c abca b c+ + ≤ + + ≤ + + −

+ +


126

� When 1b c= = , ( ) ( ) ( )2 522 1 2

* 12 27(2 1)

a aa a

a a+ +⇔ ≤ + + ≤+ +

and

( ) ( ) ( )33 2 2 4** 2 1 227

a a a a a a⇔ + ⋅ ≤ + + ≤ + −

Notice that ( ) ( ) ( )( )

( )2 5

33 2 22 1 2 42 1 2

2 27 2 1 27a a

a a a a a aa a

+ ++ ⋅ ≤ ≤ + + ≤ ≤ + −+ +

when 1a ≤

And ( ) ( ) ( )

( )

2 533 2 22 1 4 ( 2)

2 1 22 27 27 2 1

a aa a a a a a

a a+ +≤ + ≤ + + ≤ + − ≤+ +

when 1a ≥ .

Therefore for a particular inequality, we may prefer the different evaluation to get the most

effectible inequality. However remember that we should use in pair to have the best results.

i) For inequalities have equality occurs when three variable are equal and reach very near

zero when there is a variable reach 0: ( ) ( )

232 2 2 4

27ab bc ca

a b b c c a a b c abca b c+ + ≤ + + ≤ + + −

+ +

ii) For inequalities has equality occurs when three variable are equal and when apply ABC using the minimum value of abc or cabcab ++ )( cba =≤ :

( ) ( )

( )

2 52 2 2

27ab bc ca a b c

a b b c c aa b c ab bc ca+ + + +≤ + + ≤

+ + + +

iii) For inequalities has equality occurs when three variable are equal and when apply ABC

using the maximum value of abc or cabcab ++ ( )cba =≥

( ) ( )33 2 2 2 2 2 2 427

a b c a b c a b b c c a a b c abc+ + ≤ + + ≤ + + −

The reasons above are also the way for the readers find out the best evaluation for yourself when you obtain a new evaluation.

Beside that we also have a very interesting evaluation based on Vasile inequality which was mentioned in the symmetric transformation part.

Given real numbers cba ,, . Then we have the inequality: ( ) ( )23 3 3 2 2 23 a b b c c a a b c+ + ≤ + +

This inequality doesn’t evaluate directly the expression accbba 222 ++ , however we still can have its evaluation easily using the equality:

( )3 3 3 2 2 2 2 2 2 2 2 2( ) ( )a b b c c a a b c a b b c c a abc a b c a b b c c a+ + = + + + + − + + − − −

Therefore we have two good evaluations:

iv) ( )

2 2 2 2 2 2 2 2 2 22 2 2 ( ) 3 ( ) 3( )

3a b c abc a b c a b b c c a

a b b c c aa b c

+ + + + + + + ++ + ≤+ +

2 2( ) 4( ) ( ) 7( ) 3 ( )3( )

a b c a b c ab bc ca ab bc ca abc a b ca b c

+ + − + + + + + + + − + +=+ +

These inequalities look really messy but are better than all the inequalities I have given in almost cases. Also despite of its shape, ABC can handle it quite easy because of there is only

one abc inside its form.


127

Problem 12. Give cba ,, ≥ 0. Prove that: ( )2 2 2 2 2 2

2 3 3 3

125

( )a b c a b b c c aa b c a b c

+ + + ++ ≥+ + + +

(1)

Solution

(1) ⇔ ( )( )

( )( )2 2 2 2 2 2

2 3 3 3

12 35

a b c a b c ab bc ca abc ab bc caa b ca b c

+ + + + + + − − − −+ ≥+ ++ +

Apply the inequality: ( )32 2 2 427

a b b c c a a b c abc+ + ≤ + + − and we need to prove this inequality:

( )( )

( )( ) ( )32 2 2

2 3 3 3

4212 27 5a b c ab bc ca abc a b ca b c

a b ca b c

+ + + + − − + ++ + + ≥+ ++ +

(*)

The above problem can be handled using ABC because it is actually a five degrees

polynomial. We need to consider two cases:

Case 1: 1, 0b c= = , then: ( ) ( )( )

( ) ( )32

2 3

41 112 1 27* 511

a a aaaa

+ − ++⇔ + ≥++

This is true since ( )

( )( )

( )

2 2

2 2

2 1 12 11 6

1 1

a a

a a

+ +≥ � ≥+ +

and 3

3

27 ( 1) 4( 1)1 0

27( 1)a a a

a+ − ++ ≥

+

Case 2: 1b c= = , then:

(*) ( )

( )

( )( ) ( )32

2 3

42 2 1 2 212 2 27 5

22

a a a aaaa

+ + − − ++⇔ + ≥++

( )( ) ( )

( ) ( ) ( )( )

( ) ( )

2 3 2

2 3

2 223 2

2 3

12 2 4 30 33 224 1

27 22

8 1 31 32 1185 156 252 304 1 0

( 2) 27 2

a a a a

aa

a a aa a a a

a a

+ − + + +⇔ − ≥ −++

− + −⇔ ≥ ⇔ − − + − ≥+ +

Therefore the problem has been solved.

Remarks: in this problem when there is two equal variables, the most trouble case is when 1≥a .

Therefore apply the above inequality is reasonable. The value 12 still can be reduced to smaller

values, such as 10.5 … However for the following problem, we will need another convention.


Give , , 0a b c ≥ . Prove that: ( )2 2 2 2 2 2

3 3 3

7 38

( )a b c a b b c c aa b c a b c

+ + + ++ ≥+ + + +

(1)


128

Proof

(1) ⇔ ( ) ( )( )2 2 2 2 2 2

3 3 3

7 3 38

a b c a b c ab bc ca abc ab bc caa b c a b c

+ + + + + + − − − −+ ≥+ + + +

Using the inequality: 2 2 2 2 2 2 2 2 2 2

2 2 2 ( ) 3 ( ) 3( )3( )

a b c abc a b c a b b c c aa b b c c a

a b c+ + + + + + + ++ + ≤

+ +

And we need to prove that:

( ) ( ) ( ) ( ) ( ) ( )( )( )

222 2 2 2 2 2 2 2 2 2 2 2

3 3 3

7 3 3 12 38

3

a b c a b c ab bc ca abc a b c a b c a b b c c aa b c a b c a b c

+ + + + + + − + + − + + − + ++ ≥+ + + + + +

The inequality above is clearly ABC able− , we need to consider two cases :

Case 1: 0, 1c b= = , 1a ≤ (because the inequality is still symmetric when 0c = )

( ) ( ) ( )( ) ( )

( )( ) ( )

222 2 2 2 3 2 4

3 3

7 3 1 3 1 1 3 7 3 1 3 3 18 8

1 13 1 1 3 1 1

a a a a a a a a a aa aa a a a

+ + − + − + + + − −+ ≥ ⇔ + ≥+ ++ + + +

We have: ( ) ( ) ( )2 2 23 1 3 2 1 3 7 3 1 7 6

1 2 1 2 1 2a a a

a a a+ + += ⋅ ≥ � ≥

+ + + and ( )( )

3 2 4

3

3 3 10

3 1 1

a a a a

a a

+ + − − ≥+ +

Add two inequalities side by side we have: ( )

( )( )

2 3 2 4

3

7 3 1 3 3 1 7 68

1 23 1 1

a a a a aa a a

+ + + − −+ ≥ >+ + +

Case 2: 1b c= =

( ) ( ) ( ) ( ) ( ) ( )( ) ( )

( )

222 2 2

3

2 4 3 2

4 3

7 3 2 3 2 2 1 12 2 2 3 2 18

2 3 2 2

7 3 2 6 5 12 57 1

2 3( 2 2 4)

a a a a a a aa a a

a a a a aa a a a

+ + + − + − + − ++ ≥+ + +

+ − + + + +⇔ − ≥ −+ + + +

( )

( ) ( )( ) ( )

( ) ( )

2 22

32

14 1 4 8 7 1

3 2 22 3( 2) 2

a a a a

a aa a a

− + + −⇔ ≥+ ++ + + +

( ) ( ) ( ) ( )( )2 3 2 21 42 2 4 8 7 3 2 2 0a a a a a a� �⇔ − + − + + + + + ≥� �

The rest work is to prove that: ( ) ( ) ( )( )3 2 2( ) 42 2 4 8 7 3 2 2 0f a a a a a a= + − + + + + + ≥ will

be remained for the readers.

The proof end here.


129

Remarks: In the above problems, we only find the evaluation for the expressions

2 2 2a b b c c a+ + , 2 2 2ab bc ca+ + . Actually that is enough to do already, because all the rest

cyclic expressions can be represented through: 2 2 2, , ,abc ab bc ca a b c a b b c c a+ + + + + + .

We will not give a proof for the above result, however the interested readers may find out

quickly using inductive. Instead of that, we will give some equality that the readers may use

when solving problems. As normal, let assume that , ,a x y z b xy yz zx c xyz= + + = + + = :

( )( ) ( )

( )( ) ( )

( ) ( )

3 3 3 2 2 2 2

4 4 4 2 2 2 2 2 2

3 2 3 2 3 2 2 2 2 2

5 5 5 3 2 2 2 2 2 3 3

4 2 4 2 4 2 2 2 2 3 3 2

2

4 3

x y y z z x a x y y z z x b ac

x y y z z x a b x y y z z x ab bc a c

x y y z z x b x y y z z x bc a c

x y y z z x a ab c x y y z z x a b a c b

x y y z z x ab c x y y z z x abc a c b c

+ + = + + − +

+ + = − + + − + +

+ + = + + + −

+ + = − + + + − + +

+ + = − + + + − − −

Proposed Problems

Problem 14. Prove that:3 3 3

2 2 2

84

( )( )( )a b c abc

a b b c c aa b b c c a+ + + ≥

+ + ++ +, , , 0a b c∀ ≥

Problem 15. [Nguyen Anh Cuong] Prove that:

2 2 23 1

a b c ab bc cab c a a b c

+ ++ + + ≥ ++ +

, , , 0a b c∀ ≥

Problem 16. [Nguyen Anh Cuong] Prove that:

22 2 2

2 2 2 2 2 24 7

a b c ab bc cab c a a b c

+ +� �+ + + ≥� �+ + , , , 0a b c∀ ≥

Mixing Variables Method

130

§4.19. MIXING VARIABLES METHOD

Main points:

I. Introduction

II. Inequalities with three-variables

1. Unconditioned inequalities

2. MV method for conditioned inequalities

3. Trigonometric MV method in triangles

III. Using MV method with functions

IV. Three-variable inequalities involving boundary

V. SMV theorem - inequalities with four variables

VI. MV via convex function

VII. Undefined mixing variables – UMV

VIII. MV with mean value

IX. Inequality general induction (IGI)

X. Entirely mixing variables – EMV

1. EMV with the boundary at 0

2. EMV for inequalities with triangles

XI. Some special mixing variables techniques

XII. General mixing variables theorem (GMV)

XIII. Review & Proposed problems

I. INTRODUCTION

� Dear reader, in our knowledge, for most inequalities, especially symmetric or permutation inequalities, equality occurs when variables are equal. A typical example is AM GM− inequality, for example, for n = 3:

Example 1.1. Let x, y, z ≥ 0. Then 33x y z xyz+ + ≥ ⋅ .

Equality occurs if and only if x y z= = ≥ 0.

� There are many such inequalities so we usually believe that equality occurs when variables

are equal. This erroneous observation is understandable since it requires advanced mathematical

ability in order to construct a symmetric or permutation inequality so that equality occurs at a

state that all variables are not the same. Let us consider the following example.

Example 1.2. (VMO)

Let , ,x y z be real numbers such that 2 2 2 9x y z+ + = . Then ( )2 10x y z xyz+ + − ≤ .

In this case, equality occurs for (x, y, z) = (2; 2;−1) or any cyclic permutation.


131

� You might be surprised if you know that there are inequalities in which equality occurs

when all variables have distinct values. For instance

Example 1.3. (Jackgarfukel) Given , , 0 a b c ≥ . Prove that

54

a b c a b ca b b c c a

+ + ≤ + ++ + +

In this case, equality occurs when a = 3b > 0, c = 0 or any cyclic permutation. We may ask

why is (3, 1, 0)? Intuitively, we see there is something special about the fact that there is a

variable which is 0. We observe that a, b, c are non-negative, thus a variable which has value

0 is called variable which has value on the boundary (of the domain).

� Another interesting point is that for some special inequalities, equality occurs for more than

once (excluding cyclic permutation). These inequalities are beautiful and difficult. Here, we

introduce a well-known example:

Example 1.4. (Iran TST 1996) Given , , 0a b c ≥ . Prove that

( )( ) ( ) ( )2 2 2

91 1 14

ab bc caa b b c c a

� �+ + + + ≥� �+ + +� �

Equality occurs when either a = b = c > 0 or a = b > 0, c = 0 (or any cyclic permutation). If

you investigate on this example, you will see that this problem is actually consistent with the

above examples.

� In summary, in the world of inequalities, the equalities usually occur in one of the

following forms:

+ All variables are equal; we will call this “extreme value is at center”.

+ Some of variables are equal; we will call this “extreme value is symmetric.

+ One of variables is on the boundary; we will call this “extreme value is at the boundary.”

Mixing variables method (MV for short) is designed to solve such inequality by transforming

the inequality to a simple inequality after deducing the number of variables. In some cases,

after deducing the numbers of variables, we shall obtain a one-variable inequality which can

be solved by investigating on its behavior as a function.

� We now present main techniques of MV method through some specific examples. This

chapter is organized as following: in the first part, we shall consider 3-variable inequalities,

then 4-variable inequalities and in last section, we will consider the general MV for n variables.

In the last section, we begin by introducing some “classical results”, and then we obtain some

modifications and finally some general inequalities. We would like to transfer the natural idea

of how to solve problems. Afterwards, readers can solve inequalities yourself.


132

II. INEQUALITIES WITH THREE-VARIABLES

� Assume that we need to prove ( ), , 0f x y z ≥ for real numbers x, y, z satisfying some

properties. We then process the following two steps:

Step 1: (2 variables are equal)

Estimate ƒ(x, y, z) ≥ ƒ(t, t, z) for a suitable t which depends on the relationship between x, y, z,

for example, 2 2

; ;2 2

x y x yt t xy t

+ += = = ; …

Step 2: See if ƒ(t, t, z) ≥ 0.

Notice: In some cases, it is much easier to normalize variables of the inequality before applying

the above steps.

1. Unconditioned inequalities

For unconditioned inequalities, we often use the average quantities such as

2 2

; ;2 2

x y x yt t xy t

+ += = = . Let us consider following problems

Problem 2.1. Given , , 0 a b c ≥ . Prove that 33x y z xyz+ + ≥ ⋅ (1)

Proof

Method 1: The inequality (1) ⇔ ( ) 3, , 3f x y z x y z xyz= + + − ⋅ , ∀ x, y, z > 0

Step 1: Prove that: ( ) ( ), , , ,f x y z f t t z≥ where 2

x yt

+= . From: 2t xy≥ we have:

( ) ( ) ( )3 32 23 3, , , , 3 2 3 3 0f x y z f t t z x y z xyz t z t z t z xyz� �− = + + − ⋅ − + − ⋅ = − ≥� �

Step 2: Prove that ( ) 3 2, , 2 3 0f t t z t z t z= + − ⋅ ≥ .

Indeed, we have: ( ), , 0f t t z ≥ ⇔ ( ) ( ) ( )3 222 27 0 8 0t z t z t z t z+ − ≥ ⇔ − + ≥ (ans)

Conclusion: ( ) ( ), , , , 0f x y z f t t z≥ ≥

Method 2: The inequality (1) ⇔ ( ) 3, , 3 0f x y z x y z xyz= + + − ⋅ ≥ , ∀ x, y, z > 0.

Step 1: Prove that: ( ) ( ), , , ,f x y z f t t z≥ where t xy= . We have: 2t x y≤ + then:

( ) ( ) 3 3, , , , 3 2 3 2 0f x y z f t t z x y z xyz t z xyz x y t� �− = + + − ⋅ − + − ⋅ = + − ≥� �

Step 2: Prove that ( ) 3 2, , 2 3 0f t t z t z t z= + − ⋅ ≥ .

Indeed, we have: ( ), , 0f t t z ≥ ⇔ ( ) ( ) ( )3 222 27 0 8 0t z t z t z t z+ − ≥ ⇔ − + ≥ (ans)

Conclusion: ( ) ( ), , , , 0f x y z f t t z≥ ≥


133

Method 3: MV method for homogeneous inequality

Step 1: If 0x y z k+ + = > then (1) ( )3

3 27 27y yx xz zx y z xyz

k k k k k k� �⇔ + + ≥ ⇔ + + ≥ ⋅ ⋅ ⋅ � �

From that if we assume that 1x y z+ + = (*), then (1) ( ), , 1 27 0f x y z xyz⇔ = − ≥

Step 2: We also notice that after replacing x and y by 2

x yt

+= condition (*) still holds, i.e.

1t t z+ + = , thus, it is enough to consider the behavior of xyz.

According to AM GM− , 2

2

2x y

xy t+� �≤ =

� �, therefore 2xyz t z≤ ƒ(x, y, z) ≥ f(t, t, z)

Step 3: Let z = 1− 2t we have: ( ) ( ) ( ) ( ) 22, , 1 27 1 2 1 6 1 3 0f t t z t t t t= − − = + − ≥ .

Under the condition (*), equality occurs if 3 1

x y

t

=�� =�

⇔ 13

x y= = ⇔ 13

x y z= = = .

Hence, in general, equality occurs if x = y = z > 0.

Method 4:

Step 1: If 3 0xyz k= > then (1) 33y yx xz z

k k k k k k⇔ + + ≥ ⋅ ⋅ ⋅

We then assume that 1xyz = (*), thus (1) ( ), , 3 0f x y z x y z⇔ = + + − ≥

Step 2: We also notice that after replacing x and y by t xy= condition (*) still holds, i.e. . . 1t t z =

thus, it is enough to consider the behavior of x y z+ + . According to AM GM− ,

2 2x y xy t+ ≥ = , therefore 2x y z t z+ + ≥ + ƒ(x, y, z) ≥ ƒ(t, t, z).

Step 3: Let 21zt

= we have: ( ) ( ) ( )2

2 21 2 11, , 2 3 0t tf t t z t

t t− += + − = ≥ .

Under the condition (*), equality occurs if 1

x y

t

=�� =�

⇔ 1x y= = ⇔ 1x y z= = = .

Hence, in general, equality occurs if x = y = z > 0.

� Remarks:

a) Through the above example, we can prove inequality ( ), , 0f x y z ≥ by proving the

following two inequalities ( ) ( ), , , ,f x y z f t t z≥ ; ( ), , 0f t t z ≥ .

b) We can easily apply MV method to an inequality which is in a normalized form. When

considering an inequality which is not in normal form, we should choose a suitable MV

method in order to transform the inequality to the simplest form as possible.


134

� Because AM GM− (n = 3) is very simple, MV method might not be impressive to some of readers. However, you will be convinced by the effectiveness of MV method through following examples.

Problem 2.2. Let , ,a b c ∈� . Prove that ( ) ( ) ( ) ( )2 2 22 2 2 9a b c ab bc ca+ + + ≥ + +

(APMO 2004)

Proof Since the left hand side is even function with respect to a, b, c, it suffices to prove the inequality for non-negative real numbers a, b, c.

Letting ( ) ( ) ( ) ( ) ( )2 2 2, , 2 2 2 9f a b c a b c ab bc ca= + + + − + + for a, b, c ≥ 0.

We now consider the following difference:

( ) ( ) ( ) ( ) ( )2 2 22, , , , 2 4 9d f a b c f ab ab c a b a b c a b c� �= − = − + + + −� �

If we assume that c = min{a, b, c} then it is obvious that d ≥ 0 or ( ) ( ), , , ,f a b c f ab ab c≥

It remains to prove that ƒ(t, t, c) ≥ 0. By representing ƒ(t, t, c) as a quadratic trinomial of c

( ) ( ) ( )22 2 4 2, , 2 18 2 8f t t c t c tc t t= + − + − + , we have

( ) ( ) ( ) ( ) ( )2 22 2 4 2 2 4 29 2 2 8 1 2 11 32 0t t t t t t t′∆ = − + − + = − − + + ≤ ƒ(t, t, c) ≥ 0

This completes our proof. Equality occurs if a = b = c = 1.

� Remark: In symmetric inequalities, we can assume either a b c≤ ≤ or a b c≥ ≥ , and in

cyclic inequalities, we must assume either a = min{a, b, c} or a = max{a, b, c}.

Problem 2.3. Given real numbers , ,a b c . Find the minimum value of

( ) ( ) ( ) ( ) ( )4 4 4 4 4 44, ,7

f a b c a b b c c a a b c= + + + + + − + +

Solution

WLOG, assume ( ) 0a a b c+ + ≥ . Consider the difference

( ) ( ) ( ) ( ) ( ) ( )2 22 23 15, , , , 3 02 2 28 56

b c b cf a b c f a b c a a b c b c b c+ + � �− = + + + + + + − ≥� ��

Thus ( ) ( ), , , ,2 2

b c b cf a b c f a + +≥ . If 0b c+ = then ( ) 43, , 07

f a b c a= ≥

If 0b c+ ≠ , we standardize 2b c+ = . We now have ( ) ( )4 44( ,1,1) 2 1 16 2 ( )7

f a a a g a= + + − + =

It is easy to see that ( ) 0g a ≥ by investigating on the behavior of g(a). Therefore ( , , ) 0f a b c ≥

Equality occurs if and only if 0a b c= = =

Problem 2.4 (Vasile) Given , , 0a b c ≥ such that 1ab bc ca+ + = . Prove that:

2 2 2 281 1 1

2 2 2 ( )a bc b ac c ab a b c+ + ≥

+ + + + +


135

Proof

Letting ( , , )f a b c LHS RHS= − . We will prove that: ( , , ) ( , , ) ,2

b cf a b c f a t t t +≥ = .

First of all, we know 2 2 21 1

2 2a bc a t≥

+ +, therefore we only need to prove

( )2 2 2 22 2 21 1 2 ( ) 2 2 8 5 ( ) 0

2 2 2b c b c bc a a b c

b ac c ab t at+ ≥ ⇔ − + + + − + ≥

+ + +

Supposing min{ , , }a a b c= 2 2 2 2 2 22 2 8 5 ( ) 2 2 3 4 5( )( ) 0b c bc a a b c b c bc a b a c a+ + + − + = + + − + − − ≥

( , , ) ( , , )f a b c f a t t≥ . We now need to prove 2 2 2( , , ) 0 (4 7 6 ) 0f a t t a a at t≥ ⇔ − + ≥ which is true

The problem is solved. Equality occurs if and only if , 0 ( 0)a b t c t= = = ≥ and its permutations

� Remark: The above problem is an example of inequalities whose optimal value is not at the

centre. For MV method, the most important part is “mixing”, it does not matter whether

optimal value is at the centre or not. This is the distinct feature of MV method in comparison

with traditional method. In general, it is not effective to apply classical inequalities for

problems whose optimal value is not at the centre.

2. MV method for conditioned inequalities

MV method for conditioned inequalities is very different with the MV method we use in

previous section. For example, with the condition 1ab bc ca+ + = , if we want to prove

( , , ) ( , , )f a b c f a t t≥ , the variable t is not an “average quantity” of b, c but a quantity such

that 22 1at t+ = . In general:

To prove that ( , , ) 0f a b c ≥ where a, b, c satisfies ( , , ) 0g a b c = we need to prove

( , , ) ( , , )f a b c f a t t≥ where t satisfies ( , , ) 0g a t t = . We present some examples below:

Problem 2.5. Given , , 0a b c ≥ such that 1a b c+ + = . Prove that

( ) ( ) ( )2 2 2 3a b c b c a c a b+ − + + − + + − ≥

Proof

Let LHS = ( , , )f a b c . Supposing { }min , ,a a b c= . As 1a b c+ + = , so 13

a ≤ .

We have ( )2a b c a+ − ≥ . Furthermore

( ) ( ) ( ) ( )2 2 22 2b c a c a b b c a b c+ − + + − ≥ + + − − ( ) ( )2 3 8 0b c a⇔ − − ≥

Therefore LHS ( ) ( )22 1 1 3 3a a a≥ + − + − ≥ ( ) ( )23 1 4 3 0a a a⇔ − − ≥ which is true

The problem is solved completely. Equality occurs ( ) { } { }1 1 1 1 1, , , , ; , , 03 3 3 2 2

a b c⇔ = .


136

Problem 2.6. (Phan Thanh Nam) Given 3a b c+ + ≥ . Determine all real numbers k satisfying: 2 2 2 3( )( )( ) (1 )a k b k c k k+ + + ≥ + (1)

Solution

Step 1: First, we give some remarks to simplify the problem. Choose 0c = , { }3max ,2

a b k= >

which implies 0k > . Moreover, if 0k > satisfies the condition of the problem, so does every

k k′ > , since 2 2 2 2 2 2( )( )( ) ( ) ( ) ( )a k b k c k a k k k b k k k c k k k� � � � � �′ ′ ′ ′ ′ ′+ + + = + + − + + − + + −� ��

( ) ( )3 32 2 2 3 33 3( )( )( ) (1 ) (1 )a k b k c k k k k k k k′ ′ ′≥ + + + + − ≥ + + − = +

Therefore, it suffices to find the smallest real number k satisfying (1).

We now show that the following statements are equivalent, for all 1k > :

(i) (1) holds true for all real numbers a, b, c satisfying 3a b c+ + ≥ .

(ii) (1) holds true for all nonnegative real numbers a, b, c satisfying 3a b c+ + ≥ .

(iii) (1) holds true for all nonnegative real numbers a, b, c satisfying 3a b c+ + = .

(iv) (1) holds true if , 3 2a b x c x= = = − , for all ]1,0[∈x .

Indeed, the direction )()()()( iviiiiii is obvious, thus we only need to show the

reverse one. We prove )()( iiiiv

Assume that a, b, c are non-negative real numbers such that 3a b c+ + = .

By symmetry, we may suppose that bac ,≥ 0 2a b≤ + ≤ .

Letting ( ) 2 2 2 3, , ( )( )( ) (1 )f a b c a k b k c k k= + + + − + . Set 2

a bx += then ]1,0[∈x and 3 2c x= −

We have: ( ) ( ) ( )22 2( , , ) , , 2 ( ) 0

2a bf a b c f x x c a b k ab c k

� �+− = − − − + ≥� ��

since ( ) ( )2 2

2 2 02 2

a b a bk ab k� �+ +− − ≥ − ≥� ��

Besides, (iv) implies that 0),,( ≥cxxf , thus 0),,( ≥cbaf , which means that (iii) holds true.

The direction )()( iiiii is trivial for if a, b, c is non-negative then the LHS of (1) increases

as a function of a, b, c. Finally, to prove )()( iii , we may replace (a, b, c) by ( ), ,a b c

Step 2: Consider 0k > , we will find all k satisfying (iv)

We have: 2 2 2 4 3 2( , ,3 2 ) ( 1) 6 (9 6 3) 4 4 3 2 1f x x x x k x x k x x x x� �− = − + − + + − − − −� �

Letting 2 2 4 3 2( ) 6 (9 6 3) 4 4 3 2 1g k k x x k x x x x= + − + + − − − − then g is a quadratic expression in k.

Since ]1,0[∈x , it follows that 4 3 24 4 3 2 1 0x x x x− − − − < , hence g has 2 roots with opposite

signs, where the positive one is 2

33 2 1 13(11 5 )( 1)

4 12x x

y x x− + += + − + .

It is easily seen that ( ) 0g k k y≥ ⇔ ≥ (since 0k > ).

Therefore, we only need to find the maximum value of the function ( )y y x= on the domain ]1,0[∈x


137

x −3 −1 0

g ′ + 0 −

g –6

10

6 2

We have: 1 (7 5 ) 1

( ) 1 32 3(11 5 )

x xy x x

x

� �− +′ = − +� �−� ��

; 3 2( ) 0 10 27 12 1 0y x x x x′ = − + + =

This equation has a unique root in the interval [0;1] which is

0

1392 arccos41 941 41cos 0, 66318650285 3 10

x

� �� π − � � = + ≈

� �

Moreover, { }0( ) max (0), (1)y x y y> , hence y attains its maximum value on [0,1] at the point 0x .

So all real numbers 0k > satisfying (iv) are 0 0( ) 1,109926818k k y x≥ = ≈ .

Because 0 1k > , these are all values k such that (i) holds true.

Conclusion: All values of k satisfying the condition of the problems are 0 0( ) 1,109926818k k y x≥ = ≈ .

Problem 2.7. (VMO) Let x, y, z be real numbers such that 2 2 2 9x y z+ + = .

Prove that: ( )2 10x y z xyz+ + − ≤ .

Proof

Let ( ) ( ), , 2f x y z x y z xyz= + + − and 2 2

2y z

t+= .

Consider ( ) ( ) ( ) ( )2, , , , 2 2d f x y z f x t t y z t x yz t= − = + − − −

Since y + z − 2t ≤ 0 and 2 0yz t− ≤ , if x ≤ 0 then d ≤ 0. We now suppose that x = min{x, y, z}.

� If x ≤ 0 then ( ) ( ), , , ,f x y z f x t t≤ . We will to show that ƒ(x, t, t) ≤ 10

Replace29

2xt −= ( ) ( ) ( ) ( )2 21, , 2 2 2 9 9

2f x t t x x x x g x= + − − − = , [ ]3,0x ∈ − .

Considering function ( )2

2

3 5 42 2 18 2

x xg xx

′ = − −−

= 0 [ ]1 3, 0x = − ∈ −

Based on the table on the right hand side

( ) ( ) ( ) [ ], , , , 10, 3,0f x y z f x t t g x x≤ = ≤ ∀ ∈ −

( ), , 10f x y z = ⇔ x = −1, y = z = 2.

� If x > 0 y > 0, z > 0.

Without using MV method, we consider the following two cases:

If 34

x ≥ then ( ) ( ) ( ) ( )32 2 2 3 27, , 2 2 3 2 27 10

4 64f x y z x y z xyz x y z= + + − ≤ + + − = − <

If 34

x ≤ then ( ) ( ) ( )( ) ( )2 2 3 3, , 2 2 2 2 18 104 4

f x y z x y z xyz y z= + + − ≤ + + ≤ + <

The problem is solved completely. Equality occurs if 2, 1a b c= = = = − or its permutation.

Comment: The conditions of above problems are with average quantities; therefore MV method is pretty much similar to Section 1. Now, we will consider problems with special conditions


138

Problem 2.8. Given , , 0

1

a b c

ab bc ca

≥��

+ + =��. Prove that

( ) ( ) ( )2 2 291 1 14a b b c c a

+ + ≥+ + +

(Iran TST 1996)

Proof

Let LHS = ƒ(a, b, c) . Supposing t ≥ 0 is a variable such that 2 2 1t at+ = . Consider:

2 2 2 2 2

1 1 1 1 2( , , ) ( , , )

( ) (2 ) ( ) ( ) ( )d f a b c f a t t

b c t a b a c a t= − = − + + −

+ + + +

We will prove 0d ≥ by transforming d into the form of 2( )b c A− where A > 0

We have: 2 22 ( ) ( )( )t at ab bc ca a t a b a c+ = + + ⇔ + = + + . Therefore

22

2

2 ( ) ( ) 2( ) ( ) ( ) 2 ( )( )

( )( )

( )

b c t a b a c a t a b a c a b a c

b ca b a c

a b a c

+ − = + + + − + = + + + − + +

−= + − + =+ + +

So 2 2 2 2

(2 )(2 ) 1 1 2( )( )( ) (2 ) ( ) ( )

t b c t b cd

a b a cb c t a b a c

− − + += + + −+ ++ + +

2 2

2 22 2 2

22 2 2 2 2

( ) (2 ) ( )( ) ( )( ) ( ) (2 )

1 2( )

( ) ( ) ( ) ( ) (2 )

b c t b c b c

a b a ca b a c b c t

t b cb c

a b a c a b a c b c t

− − + + −= ++ ++ + + +

+ +� �= − − + + + + + +� �

Now, let us assume min{ , , }a a b c= , thus a t≤ and 22 ( )t b c a b a c+ + ≤ + + + ,

2 2 2 2( ) ( ) ( ) (2 )a b a c b c t+ + ≤ + . Therefore 0d ≥ or equivalently ( , , ) ( , , )f a b c f a t t≥

On the other hand, replace 21

2tat

−= into ( , , )f a t t and after transformation, we get

( ) ( ) ( )( )

22 2

2 21 1 3 9 9, ,

4 44 1t tf t t ct t

− −= + ≥+

. The problem is solved.

Equality occurs if and only if 13

a b c= = = or 0, 1a b c= = = and its permutation

Problem 2.9 [Le Trung Kien, Vo Quoc Ba Can]

Given , , 0a b c ≥ such that 6 9ab bc ca abc+ + + = . Prove that 3 6a b c abc+ + + ≥

Proof

Letting ( , , ) 3f a b c a b c abc= + + + . Supposing 2 22 6 9at t at+ + = (0 3)t≤ ≤


139

Consider the difference: 2( , , ) ( , , ) ( 2 ) 3 ( )d f a b c f a t t b c t a bc t= − = + − + −

With the given conditions, to transform ( 2 )b c t+ − , 2( )bc t− into 2( )A b c− is really complicated.

We will deal with this in a more “sophisticated” way

The condition 2 2 26 2 6 ( 2 )6 1

aab bc ca abc at t at b c t t bca

+ + + = + + ⇔ + − = −+

Notice that if 2

2 0

20

b c t b ct bc

t bc

+ − <� +� ⇔ < <�− <��

. This can never happen. Therefore 2

2 0

0

b c t

t bc

+ − ≥��

− ≥��

So 2

2 3( 2 ) 3 ( ) ( 2 ) 1

6 1a

d b c t a bc t b c ta

� �= + − + − = + − − +� �

. Now, let us assume min{ , , }a a b c= ,

then 1a ≤ which implies 23

1 06 1

aa

− ≥+

. Thus 0d ≥ or equivalently ( , , ) ( , , )f a b c f a t t≥

Replace 2

2

92 6

ta

t t−=+

and after transformation, we have ( )2

2

3(3 )( 1)( 1), , 6 6

2 6t t t

f a t tt t

− + −= + ≥+

The problem is solved completely.

Equality occurs if 1a b c= = = or 0, 3a b c= = = or its permutation

3. Trigonometric MV method in triangles

Trigonometric inequalities related to triangles are also 3-variable inequalities. We present few

examples to illustrate the power of MV method for this kind of inequalities:

Problem 2.10. Given ∆ABC. Find the minimum value of ( ) ( ) ( )2 2 21 os 1 os 1 osc A c B c C+ + +

Proof

Supposing min{A,B,C}C = 03

C π≤ ≤ (*). Letting ( ) ( ) ( ) ( )2 2 2, , 1 cos 1 cos 1 cosf A B C A B C= + + +

We will prove that: ( ) ( ), , , ,2 2

A B A Bf A B C f C+ +≥ (1). Indeed, we have:

(1) ( ) ( ) ( )22 2 21 os 1 os 1 os

2A Bc A c B c +⇔ + + ≥ + ( )[ ]2sin 6cos cos 1 0

2A B C A B−⇔ − − − ≥ (2)

Because of (*) we have: ( )6cos cos 1 3 1 1 0C A B− − − ≥ − − > (2) is true (1) is also true,

or equivalently ( ) ( ) ( ), , , ,2 2

A B A Bf A B C f C g C+ +≥ = = ( ) ( )2 21 3 os 1 os4

c C c C− +

It is easy to see that ( ) 12564

g C ≥ . From (1) ( ) 125, ,64

f A B C ≥

Equality occurs ABC⇔ ∆ is an equilateral triangle. Thus 125min64

f =


140

Problem 2.11. Given a non-obtuse triangle ABC .

Find the minimum value of sin sin sinos os os

A B CPc A c B c C

+ +=+ +

Solution

Supposing { }ax , ,A m A B C= , we have 2 3

Aπ π≥ ≥ . Letting [ ]os ; 0,12

B Cx c x−= ∈

( )sin 2 os

2os 2sin

2

AA c xP f x

Ac A x

+= =

+ ( )

( )2

3os2 0

os 2sin2

Acf x

Ac A x′ = ≤

+ ( )f x is decreasing in [ ]0,1

( ) ( )sin 2 os

2

os 2sin2

AA cf x g A

Ac A

+≥ =

+. We also have ( )

( )2

3sin 12 0

os 2sin2

A

g AAc A

−′ = ≤

+

( )g A is decreasing ),3 2π π��

( ) ( ) 212 2

g A g π≥ = + 212

P ≥ +

Equality occurs if and only if: ;2 4

A B Cπ π= = = or its permutations. Thus 2min 12

P = +

Problem 2.12. Given a non-obtuse triangle ABC . Prove that:

( ) ( ) ( )2 2 2sin sin sin sin sin sin 9sin sin sin 4A B B C A C

C A B+ + ≥

Proof

WLOG, we assume that 2 3

Aπ π≥ ≥ . We can rewrite the above inequality in the form

( ) ( )2 2 2 29, , 2 sin sin sin4

f A B C A B C≥ + + +

where ( ) sin sin sin sin sin sin, ,sin sin sinA B A C B Cf A B C

C B A= + +

Consider the difference:

2 2 2sin 4sin sin 12 2( , , ) , ,sin sin2 2 sin 2

B C AAB C B Cd f A B C f A

B CA

− � �+ +� � = − = −

� � � �

Since 2 3

Aπ π> ≥ , we have

2 2

44sin sin

2 16 in 1sin sin 2

AAAs

B C≥ ≥ . Hence 0d ≥

We also notice that 2 2 2sin sin 2 os2AB C c+ ≤ , so we only need to prove:

( ) ( )2 2 2 29, , 2 sin sin sin2 2 4

B C B Cf A A B C+ + ≥ + + + ( ) ( )2cos os 1 2cos 1 0A c A A⇔ + − ≥

This is true because 2 3

Aπ π≥ ≥ . The inequality is then solved.

Equality occurs ⇔ ∆ABC is an equilateral triangle or an isosceles right triangle.

Comment: If we use the formula 1sin sin

sin cot cotB C

A B C=

+ and cot .cot 1B C =� then the above

inequality is equivalent to (Iran 1996): ( )( ) ( ) ( )2 2 2

91 1 14

ab bc caa b b c c a

� �+ + + + ≥� �+ + +� �


141

III. USING MV METHOD WITH FUNCTIONS

In this section, we will study MV method with functions; this is an important technique in

MV method. In section II. in order to prove ƒ(x, y, z) ≥ ƒ(t, t, z) we consider the expression

( ) ( ), , , ,d f x y z f t t z= − then we show that d ≥ 0. It is a suitable if ƒ is a simple polynomial or

polynomial fraction. However with more complicated function, such as

( ), , k k kf x y z x y z= + + (for 0k > ) we need to evaluate intermediate quantity by

investigating the behavior of the function. In details, assume that we need to prove ƒ(x, y, z) ≥

ƒ(x, t, t) where 2

y zt

+= , let us consider function ( ) ( ), ,g s f x t s t s= + − for s ≥ 0. We then

prove that g is increasing for ∀s ≥ 0, thus g(s) ≥ g(0).

We would like to emphasize that this is a difficult and sophisticated technique in MV method.

The following examples illustrates the beauty as well as the power of MV method

Problem 3.1. Given k ≥ 0 and a, b, c ≥ 0. Prove that:

( ) ( ) ( ) { }3min 2,2

k k k

ka b c

b c c a a b+ + ≥

+ + + (*)

Proof

Step 1: It is enough to show the inequality under the case 3 ln 32 1ln 22 k

k= ⇔ = −

We leave it as a food for thought for readers to find out the reason.

Notice: For ln 3 1ln 2

k = − equality occurs for a = b = c > 0 or , 0, 0, 0

a b c

b c a

c a b

= =�� = =�� = =�

Step 2: Without lost of generality, we assume a + b + c = 1 and b ≥ c ≥ a.

Let 2

b ct += and 2

b cm −= , it follows that b = t + m, c = t − m, a = 1− 2t. Therefore:

(*) ⇔ ( ) ( ) ( ) ( )1 2 22 1 1

k k kt t m t mf mt t m m t

− + −= + + ≥− − + −

where ln 3 1ln 2

k = −

Since c ≥ a, we have 3t − 1 ≥ m ≥ 0, and 1 ≥ b + c = 2t thus 1 12 3

t≥ ≥ .

Let us consider function ƒ(m) where m∈ [0, 3t − 1] and 1 1,3 2

t � �∈� ��

is a constant.

We have ( ) ( )( )

( )( )

1 1

1 11 1

k k

k kk t m k t mf m

t m m t

− −

+ ++ −′ = −

− − + −

( )( )

( )( )

1 1

1 10

1 1

k k

k kk t m k t m

t m m t

− −

+ ++ −≥ ⇔ ≥

− − + −

( ) ( ) ( )[ ] ( ) ( )[ ]1ln ln ln 1 ln 1 01

kg m t m t m t m t mk

+⇔ = − − + − − − − − + ≥−


142

Then: ( ) ( ) ( )11 1 1 1 01 1 1

kg mt m t m k t m t m

+′ = + + + ≥− + − − − − +

⇔

( ) ( )( )

( ) ( )2 12 1 0

1 1 1tt k

kt m t m t m t m−− ++ ⋅ ≥

−− + − − − + ( )2 2 2 2

1 1 01 1

t k tkt m t m

− + −⇔ + ⋅ ≥−− − −

(1)

Since ln 3 1ln 2

k = − , 1 21

kk

+ ≥−

. Therefore, it is enough to prove that

( )( )

( ) 2 3 2 22 2 2 2

2 1 0 4 3 3 2 01

tt u m t t t tm mt m t m

−− + ≥ ⇔ = − + − + − ≥− − −

Now ( ) ( )2 3 2 0u m t m′ = − < ∀ 1 1,3 2

t � �∈� ��

u(m) ≥ u(3t − 1) = 2(3t − 1)(2t − 1)2 ≥ 0.

(1) holds g′(m) ≥ 0 g(m) is increasing g(m) ≥ g(0) = 0 ƒ′(m) ≥ 0

ƒ(m) is increasing ƒ(m) ≥ ƒ(0) ( ) ( )1 2 22 1

k kt tt t

−= +−

.

Step 3: We shall prove that ( ) ( ) ( ) ( )1 20 2 22 1

k kt tf h tt t

−= = + ≥−

, ∀ 10,3

t � �∈� ��

( )( )

( ) ( ) ( )[ ]11 11 2

1 11 22 0 2 1 1 2

21

kk kk kk k k

tkt kh t t t ttt

−− −++ +

−′ = − ⋅ ≤ ⇔ ≤ − −−

(2)

In the last inequality, the left hand side is an increasing function with respect to t and the

right hand side is a decreasing function with respect to t, and since 13

t ≤ then it is sufficient

to show that: ( ) ( ) ( ) 121 1 1 22 1 1

3 3 3

kkk

−+ � �≤ − −

� �� .

It is easy to see that the above inequality holds, thus h(t) is decreasing.

Therefore ( ) ( )1 23

h t h≥ = which completes our proof. Comparing 1 1,3 2

t � �∈� ��

and 10,3

t � �∈� ��

Remarks: To see the beauty of above inequality, we consider some special cases

a) For k = 1, we get Nesbit inequality: 32

a b cb c c a a b

+ + ≥+ + +

There are 12 ways to prove the inequality, e.g.

3a b c a b c a b c a b cb c c a a b b c c a a b

+ + + + + ++ + + = + ++ + + + + +

( ) ( ) ( )( ) ( ) ( )

9 91 1 12

a b c a b cb c c a a b b c c a a b

= + + + + ≥ + + =+ + + + + + + +

b) For 12

k = , we get: 2a b cb c c a a b

+ + ≥+ + +

Here is a solution using AM – GM inequality:

( )

( )22 2 22cyc cyc cyc

a b ca a ab c a b c a b ca b c

+ += ≥ = =+ + + + ++

� � �


143

c) For 23

k ≥ , we have: ( ) ( ) ( ) 32

k k k

ka b c

b c c a a b+ + ≥

+ + +

This is a nice inequality although constant 23

k = is not the best. Here is a solution

( ) ( )2233 2 33

2 2 2b c b c b c a aa b c a a

b c a b c+ + ++ + = + + ≥ ⋅ ≥

+ + + ( )

23

23

3

2cyc

ab c

≥+�

Problem 3.2. Let k > 0, a, b, c ≥ 0 and a + b + c = 3. Prove that:

( ) ( ) ( ) ( ) 23max 3,2

kk k kab bc ca

� �+ + ≤ � �

� � (1)

Proof

Without lost of generality we may assume that b ≥ c (we shall choose a = min{a, b, c} or a = max{a, b, c} in a suitable way).

Let 2

b ct += and 2

b cm −= it follows that b = t + m, c = t −m. Then we can rewrite (1) as following:

( ) ( ) ( ) ( ) ( )22 2 3max 3,

2

kkk kkf m a t m t m t m� �� = + + − + − ≤ � ��

Now consider ƒ(m) on m∈ [0, t]. After some manipulations, we get:

( ) ( ) ( ) ( ) 11 1 2 22kk kkf m ka t m t m km t m

−− −� �′ = + − − − −� �

( ) ( ) ( ) ( )1 10 2 0k kkf m g m a t m t m m− −� �′ ≥ ⇔ = − − + − ≥� �

It is enough to consider that case k > 1 (for k ≤ 1 the inequality is trivial).

Since ( ) ( ) ( ) ( )11 0k kkg m a k k t m t m− − −� �′′ = − − − + >� � , g′(m) is increasing, it follows that g′(m) = 0

has at most one solution in (0, t). Because g(0) = 0, g(t) = + ∞ there are two possible cases ( ) 0g m > ,∀m∈(0; t] or ( ) 0g m = − + ,∀m∈ [0, t].

⇔ ƒ′(m) > 0 ,∀m∈(0; t] or ƒ′(m) = −0+ ,∀m∈ [0, t]

⇔ ƒ(m) goes up or ƒ(m) goes down and then goes up.

In both cases, the maximum value occurs in the boundary, this means ( ) ( ) ( ){ }max 0 ,f m f f t≤

For m = t c = 0 ( ) ( ) ( ) ( )2 232 2

k kk a bf t ab += ≤ =

For m = 0 b = c = t ( ) ( ) ( )2 20 2 2 3 2 kk k k k kf t a t t t t h t= + = − + =

We have: ( ) ( ) ( )1 1 2 14 3 2 2 3 2 2k kk k kh t k t t k t t kt− − −′ = − − + − +

( ) ( ) ( ) ( )1

13 2 3 20 2 1 0 2 1 0k k

k kt th t u x x xt t

−−− −′ ≥ ⇔ − + + ≥ ⇔ = − + ≥ for 3 2tx

t−=

We get: ( ) ( )[ ] 22 1 ku x kx k x −′ = − − . It follows from that fact that u′(x) has at most one solution

in +� that u(x) has at most two solutions in +

� , one of which is x = 1. By the way, we now


144

suppose that a = min{a, b, c}. It is enough to consider the case t ≥ 1 or equivalently x ≤ 1.

Since u(x) has at most one solution in (0, 1) then h′(t) has at most one solution in ( )31,2

.

Notice that h′(1) = 0, ( )3 02

h′ > . Therefore h(t) is increasing in 31,2

� ��

or h(t) has a form (−0+)

for 31,2

t � �∈� ��

. In both cases, the maximal value of h(t) occurs in the boundary, this means that

( ) ( ) ( ){ } ( )23 3max 1 , max 3,2 2

k

h t f f� �

≤ = � ��

� Remarks: We do not assume a = min{a, b, c} from beginning to point out that by applying

mix variables method, we do not need to assume any orders between the variables.

Furthermore, after proving

we can continue in a different way as follows.

For a, b, c fixed, consider the following sequence defined by:

( ) ( )0 0 0, , , ,a b c a b c= ; ( ) 2 2 2 2 2 2 2 22 1 2 1 2 1 2 2, , , ,

2 2n n n n

n n n n

b c b ca b c a − − − −

− − − −

+ +� �= � �

n +∀ ∈�

and ( ) 2 1 2 1 2 1 2 12 2 2 2 1, , , ,

2 2n n n n

n n n n

a b a ba b c c− − − −

−

+ +� �= � �

n +∀ ∈� . Then we have:

( ) ( ) ( )23, , max , , , ,

2

k

n n nf a b c f a b c n +� �≤ ∀ ∈� �

� ��

Since a + b + c = 3 then sequences { } { } { }, ,n n na b c converge to 1, it follows that:

( ) ( ) ( ) ( )2 23 3, , max , 1,1,1 max ,32 2

k k

f a b c f� � � �

≤ =� � � ��

(ans)

� Comments: We shall generalize the above technique to obtain some well-known MV methods,

namely, strongly mixing variables (SMV for short) and undefined mixing variables (UMV for

short) which will be introduced in the next section. By using the continuity of function, we

obtain more generalized methods than SMV and UMV.

On the other hand, after having (*), we have some special ways to obtain the solution; this

will be introduced in the section on 4-variable problems. In the section on 3-variable

inequality, we use only simple approach.

Here is another example, in which we use geometric mean.

(*)

( ) ( ) ( )23, , max , , ,

2

k

f a b c f a t t� �

≤ � ��

for 2

b ct +=

( ) ( ) ( )23, , max , , ,

2

k

f a b c f t t c� �

≤ � ��

for 2

a bt +=


145

Problem 3.3. (Pham Kim Hung) Given a, b, c > 0 such that abc = 1. Prove that:

a) ( ) ( ) ( ) ( ) 42 2 281 1 1 1 8a b c a b c+ + + ≤ + +

b) ( ) ( ) ( ) ( )63 3 364 1 1 1a b c a b c+ + + ≤ + +

Proof

a) Letting ( ) ( ) ( ) ( ) ( )4 2 2 2, , 8 81 1 1 1f a b c a b c a b c= + + − + + + . We then assume that a ≥ b.

Consider function ( ) ( ), ,bg t f ta ct

= where ,1bta

� �∈ � �� . We have:

( ) ( ) ( ) ( )3

22 2

32 81 1b b b bg t a ta c a ta ct tt t

� � � �′ = − + + − − + + � � � �

Since ,1bta

� �∈ � �� it follows that g′(t) ≥ 0 provided that: ( ) ( )3 232 81 1d c d c+ ≥ + for bd ta

t= +

Indeed: ( ) ( ) ( ) ( )3 32 2 4 2 2 232 32 2 3 32 3 3 81 1d c d d dc c d d c c d c+ ≥ + + ≥ + > + since 2 4d c ≥ .

Thus g′(t) ≥ 0 for ,1bta

� �∈ � �� therefore g(t) is increasing in ,1b

a� ��

( )1 bg ga

� �≥ � �

,

this means that ( ) ( ), , , ,f a b c f s s c≥ where s ab= .

Finally, we prove that ƒ(s, s, c) ≥ 0 for s2c = 1. Let 1sc

= , we get:

( ) ( ) ( )4 2

21 1 2 1, , , , 8 81 1 1f s s c f c c ccc c c

� � � �= = + − + + � � � �

( )2 9 9 5 35 4 3 22 2 2 21 8 16 24 96 87 78 99 120 21 94 47 0c c c c c c c c c c c

c� �−= + + + + + + + − + + ≥ � �

(ans)

This completes the proof. Equality occurs if 1a b c= = = .

b) Define ( ) ( ) ( )( )( )6 3 3 3, , 64 1 1 1f a b c a b c a b c= + + − + + +

Suppose cba ≥≥ . Consider the function ( ) ( , , )bg t f ta ct

= where ,1bta

� �∈ � �� . We have:

( )5 22 2 3

2 2 2( ) 6 192 (1 )b b b bg t a ta c a t a ab c

tt t t� �� ′ = − + + − − + + +

� � � ��

Since ,1bta

� �∈ � �� , ( ) 0g t′ ≥ if the inequality ( )5 2

2 2 32

32 (1 )b bta c t a ab ct t

� �+ + ≥ + + + � �

holds true.


146

Setting 2

2 22

bd t a abt

= + + , we have: ( ) ( )5 4

3b bta c ta ct t

+ + ≥ + +

( ) ( )2

2 33 2 2 2 3( 6) 72 32 1bcd ab tac d d d ct

= + + + ⋅ ≥ + ≥ > + (since 1c ≤ )

Hence g′(t) ≥ 0 for ,1bta

� �∈ � �� , which implies that g is an increasing function on ,1b

a� ��

.

In particular, (1) bg ga

� �≥ � �

, which means ( , , ) ( , , )f a b c f s s c≥ where abs = .

Finally, we show that ƒ(s, s, c) ≥ 0 where s2c = 1. Substituting 1sc

= , we obtain:

6 231 1 2 1( , , ) , , 64 1 (1 )f s s c f c c c

c c c c c� � � � � �= = + − + + � � � � � �

6 4 3 6412 4 32 112 0c c c c c cc c

= + − + + + > since 1c ≤ . The proof is completed.

Problem 3.4. (Phan Thanh Viet) Given a, b, c ≥ 0, a + b + c = 1. Determine the greatest

positive real number k satisfying: 3)()()( 222 ≤−++−++−+ bakcackbcbka (1)

Solution

� Step 1: Let 1, 0a b c= = = . Then (1) implies that 312

k ≤ − .

We now show that inequality (1) holds true for 312

k = − .

Define 2 2 2( , , ) ( ) ( ) ( )f a b c a k b c b k c a c k a b= + − + + − + + −

Without loss of generality, we may suppose that a b c≥ ≥ .

We prove ( , , ) ( , , )f a b c f a t t≤ (2) where 2

b ct += . Let b t s= + , c t s= − .

Since 1a b c+ + = , 1 2a t= − . Then the conditions 0a b c≥ ≥ ≥ implies 1 3 4 0t s s≥ + ≥ ≥ .

Consider 2 2 2( , , ) ( ) 1 2 4 (3 1) (3 1)f a b c g s t ks t s k t s t s k t s= = − + + + + − − + − + + − ,

where 10;4

s � �∈� ��

and 10;3

t � �∈� ��

. Note that (2) is true if we have ( ) (0)g s g≤ , so it suffices to

show that g(s) is an decreasing function. We have:


147

2 2 2

4 1 2 6 2 1 2 6 2( )

1 2 4 2 (3 1) 2 (3 1)

ks k kt ks k kt ksg s

t ks t s k t s t s k t s

+ − + − − + +′ = + +− + + + − − − + + −

Since 2 21 2 4 (3 1) (1 3 )(3 3 1 ) 0t ks t s k t s s t kt ks k− + ≥ − + + − ⇔ + − + + − ≥ , to obtain 0)(' ≤sg

we only need to show that 2 2

1 2 6 2 1 2 6 100

2 (3 1) 2 (3 1)

k kt ks k kt ks

t s k t s t s k t s

+ − + − − + ++ ≤+ + − − − + + −

2 2 2 2(1 2 6 10 ) ( (3 1) ) (1 2 6 2 ) ( (3 1) )k kt ks t s k t s k kt ks t s k t s⇔ + − − + + − − ≥ + − + − + + −

2 2 2 2 3 3 2 2 2 2

3 3 3 2 3 3 3 3 3 3 2 3 2

( ) 1 24 30 72 144 144 432 6 32 8 40

16 16 80 48 s 96 432 144 240 0

h s k t kt k t k ts k t k t k k s ks k s

k s k k s k k ts k t k t s k ts

⇔ = − − + + + − + − − +

+ − + + − + + − ≥

We have: 2 2 2 3 3 3 2 3 3 2 3( ) 144 32 8 80 16 160 144 96 144 480h s k t k k k s k k s k s k t k t k ts′ = − − + + + + − + −

2 3 3 3( ) 80 160 288 480 0h s k k k s k t′′ = + + − ≥ 10;4

s � �∀ ∈� ��

(since 10;3

t � �∈� ��

)

Hence ( ) ( )1 04

h s h′ ′≤ < , which implies ( ) ( )1 04

h s h≥ > . We may now conclude that ( ) 0g s′ ≤

It follows that ( , , ) ( ) (0) ( , , )f a b c g s g f a t t= ≤ = where 2

b ct += .

� Step 2: It remains to prove the inequality for the case b c= :

( )22 22(1 ) (1 3 ) 3 2(1 ) (1 3 ) 3a a k a a k a a+ − + − ≤ ⇔ − + − ≤ −

2 2 2 2(1 3 ) (1 3 ) (1 3 ) (1 3 ) 1 0k a a a k a� �⇔ − ≤ − ⇔ − + − ≤� �

The last inequality is obviously true since 2 2 1(1 3 ) (1 3)ak

+ ≤ + = .

Therefore, the greatest real number k satisfying the condition of the problem is 312

k = − .

Comment: From understanding the method to applying the method skillfully is a long way.

The most important thing is that you need to be willing to deal with the problem to the very

end, do not stop when you confront complicated calculations. Successes will make you more

confident. We now present a problem in which the solution might frighten some of you,

however we hope you will be calm to see hidden beauty of the problem

Problem 3.5. Given , , 0

3

a b c

a b c

≥��

+ + =��. Find the maximum value of S =

2 2 23 3 3ab bc ca

c a b+ +

+ + +

Solution

WLOG, supposing a ≥ b ≥ c.


148

Put a = s + t, b = s − t then we can rewrite S as: ( ) ( )( )

( )( )

2 2

2 2 233 3

c s t c s t s tf tcs t s t

− + −= + +++ + + −

We now investigate ƒ(t) for t ∈ [0, s − c]. We have:

( )( )

( )( ) ( )

( )( )

2 2 2 2

2 2 2 2 22 2

2 2 233 33 3

c s t c s tc c tf tcs t s ts t s t

− −−′ = − + + −++ + + −� � � �+ + + −� � � �

( ) ( ) ( )2 2

2 2 284 2 , 0,

3cst s t u vcst t t s c

uv u v c− += + − ∀ ∈ −

+ where ( ) ( )2 23 , 3u s t v s t= + + = + − .

If ƒ′(t) < 0, ∀t ∈(0, s – c) (we shall prove it later), then:

( ) ( ) ( )( )

( )2 2

2 2 2 22 3 220

3 3 3 3 3 2

s scs s sf t f g ss c s s

−≤ = + = + =+ + + + −

(1)

Consider g(s) for 31;2

s � �∈� ��

. We see that:

( )( ) ( )

( ) ( ) ( )( ) ( )

22 22 2

2 2 2 22 22 2

108 3 4 1 3 624 12 18 24 6

33 3 2 3 3 2 3

s s s s ss s s sg sss s s

− + − − − +− − −′ = + =� � � �++ − + − +� � � �

Obviously 2 3 4 0s s− + > and 2 33 3 33 33 62 2

s s s s� �� − +− − + = − + � ��

thus g′(s) is positive in (1, s0) and negative in ( )03;2

s for 033 3 1,372281323...

2s −= =

Hence for every 31;2

s � �∈� ��

we always have: ( ) ( )011 33 45

24g s g s −≤ = (2)

In (1) and (2), equality occurs if t = 0 and 0s s= , or equivalently if 0a b s= = and 03 2c s= − .

So, the maximum value is 11 33 45 0.757924546...24

− = when 33 3 1,372281323...2

a b −= = = ,

6 33 0.255437353...c = − =

Finally we prove ƒ′(t) < 0, ∀t ∈ (0, s − c). We shall prove

for t ∈ (0, s − c) that: 2

4 13

csuv c

<+

(3) and ( )( )2 2

2 2 28 1

3cs s t u v

u v c− + ≤

+ (4)

Prove (3): It follows from 2 1c s+ = and s > 1 that 1cs < . Moreover:

( ) ( )2 2 23 4, 3 3u s t v s t c= + + > = + − > + which yields (3).


149

Prove (4): Using AM – GM inequality we have:

( ) ( ) ( )2 22 22 2 2 23 3 16u v s t s t s t� � � �= + + + − ≥ −� � � �

and ( ) ( ) ( ) ( )32 2 2

2 2 2 2 4 3 32 3 4 3 33

cs s t ccs u v c cs s t c � �+ + + + ++ + = + + + ≤ � �

Using c = 3− 2s, together with t ≤ s − c = 3s − 3, we yield:

( ) ( ) ( ) ( ) ( )2 22 2 2 24 3 3 4 3 2 6 3 3 3 2 12 6 1 2 12cs s t c s s s s s s s+ + + + + ≤ − + + + − + − = + − − ≤

then 2cs(u + v)(3 + c2) < 43.

Thus: ( ) ( ) ( ) ( )2 2 22 2 3

2 2 2 2 2 4 2 28 2 3 1 4 14 4

3 4 3 3cs s t u v cs u v cs t

u v u v c c c− + + +−= ⋅ ⋅ ≤ ⋅ ⋅ =

+ + +

It follows from (3) and (4) that ƒ′(t) < 0, ∀t ∈ (0, s − c).

Hence 11 33 45Max 0.757924546...24

S −= =

� Remark: We can also apply the above technique to the following problem:

Problem 3.6. Let a, b, c > 0 and a + b + c = 3. Find the maximum value of:

( )2 2 2

, , bc ca abS a b ca k b k c k

= + ++ + +

where k ≥ 3 is a constant.

Problem 3.7. Let a, b, c > 0; a + b + c = 3 and ( )2 2 2

, , bc ca abS a b ca k b k c k

= + ++ + +

Find all k such that ( ) ( ) 3, , 1,1,11

S a b c Sk

≤ =+

Firstly, since ( ) ( )3 31,1,1 , , 02 2

S S≥ then k ≥ 3. Using similar approach as in 3.4 we then need

to consider the case a = b ≥ c and ( ){ }max 0 0 03max , , ,3

1S S s s s

k= −

+

where s0 is the greatest solution of the equation 6s3 + (7k − 9)s2

− 18ks + k2 + 9k.

Thus we must find all k ≥ 3 satisfying: ( )0 0 03, , 3

1S s s s

k− ≤

+

The possible value of k is [3, k0] where k0 = 3,2690313... (solution of an equation with high

degree). We also notice that in the most of inequalities concerning to find the best value of a

constant (which is complicated), we must use MV method.


150

IV. THREE-VARIABLE INEQUALITIES INVOLVING BOUNDARY

In the previous section, "mixing variables" means "two variables are equal", and now, in this

section, "mixing variables" should be understood as "moving one variable to the boundary".

For example, if we want to prove ƒ(x, y, z) ≥ 0 for x, y, z ≥ 0, we hope that ƒ(x, y, z) ≥ ƒ(0, s,

t), for suitable s, t corresponding to a, b, c. Finally, it is enough to show that ƒ(0, s, t) ≥ 0.

We now begin with Schur inequality.

Problem 4.1. Let a, b, c ≥ 0. Prove that ( ) ( ) ( )3 3 3 2 2 23a b c abc a b c b c a c a b+ + + ≥ + + + + +

(Schur inequality)

Proof

In II., we have already proved it by using MV method when two variables are equal. We also

observe that equality occurs either a = b = c or a = b, c = 0 (and any cyclic permutation).

Let ( ) ( ) ( ) ( )3 3 3 2 2 2, , 3f a b c a b c abc a b c b c a c a b= + + + − + − + − +

We hope that ƒ(a, b, c) ≥ ƒ(0, a + b, c). Now we consider:

d = ƒ(a, b, c) – ƒ(0, a + b, c) = ab(5c – 4a – 4b)

We then find that it is impossible to get d ≥ 0 for every a, b, c.

Unfortunately it is not true! But why not d ≥ 0. We now observe that ƒ(a, b, c) is decreasing

if two variables are touching (which occurred in II.), now if we replace (a, b, c) by ( )0, ,a b c+

variables are moving far from each other. We now replace (a, b, c) by ( )0, ,2 2a ab c+ + and

consider: ( ) ( ) ( ) ( ), , 0, , 2 22 2aa ad f a b c f b c a a b c a c b= − + + = + − + −

We now can assume that da ≥ 0. Indeed, by symmetry, we may assume that { }max , ,a a b cd d d d=

So, if da < 0 then ( ) ( ) ( )2 2 20 2 2 2a b cd d d abc b c a c a b a b c> = + − + − + − , a contradiction!

Thus da ≥ 0 then ƒ(a, b, c) ≥ ƒ(0, s, t) where ,2 2a as b t c= + = + . Finally, we see that

( ) ( ) ( ) 23 3 2 20, , 0f s t t s t s ts t s t s= + − − = + − ≥ this completes our proof.


151

Problem 4.2. (Hojoo Lee) Given , , 0

1

a b c

ab bc ca

≥��

+ + =�� (*). Prove that 51 1 1

2a b b c c a+ + ≥

+ + +

Proof

We notice that equality occurs for a = b = 1, c = 0 and any cyclic permutation.

If c = 0, then we must prove: A = 51 1 12a b a b

+ + ≥+

, for ab = 1.

Let s = a + b 2 2s ab≥ = . Then A = ( ) 3 3 2 51 1 124 4 4 4 2s s ss

s s s⋅+ = + + ≥ ⋅ + =

Let ƒ(a, b, c) 1 1 1a b b c c a

= + ++ + +

. We hope that ( ) ( )1, , , , 0f a b c f a ba b

≥ ++

(due to ab + bc + ca = 1). Consider: ( ) ( )1, , , , 0f a b c f a ba b

− ++

1 1 1 1 11 1 1

a ba b ab ab a b a ba b

a ba b a b

� � � �= + + − + + + + − − + + ++ + ++ + � ��

( )2 2 21 1 11

1 1 1a b a b= + − −

+ + + + ( ) ( )

( )( )

22 2

22 2

221 1 1

a ba ba b a b

+ ++ += −+ + + +

[ ] ( ) ( ) ( ) ( )( ) ( ) ( )

2 22 2 2 2

22 2

2 1 1 1 2

1 1 1

a b a b a b a b

a b a b

� � � �+ + + + − + + + +� � � �=� �+ + + +� �

( )( ) ( ) ( )

( ) ( )( ) ( ) ( )

2 22 2

2 22 2 2 2

2 2 2 1

1 1 1 1 1 1

ab a b a b ab ab ab a b

a b a b a b a b

� � � �− + + − − +� � � �= =� � � �+ + + + + + + +� � � �

Then d ≥ 0 provided 2(1 − ab) ≥ ab(a + b)2. Therefore, assume that c = max{a, b, c},

we get: 2(1− ab) = 2c(a+ b) ≥ (a+ b)2 ≥ ab(a+ b)2 d ≥ 0 (ans)

� Remark: Problem 4.2 is interesting but it is a corollary of a well-know inequality, namely

Iran TST 1996. Indeed, since ab + bc + ca = 1 then it follows from Iran TST 1996 that

( ) ( ) ( ) ( )( )

( ) ( ) ( )2

2 2 24 9 251 1 1 1 1 1 4

4 4a b c

a b b c c a a b b c c aa b b c c a

+ ++ + = + + + ≥ + =+ + + + + ++ + +

due to a + b + c = (a + b + c)(ab + bc + ca) ≥ (a + b)(b + c)(c + a).

It is natural to raise a question that whether Inequality Iran TST 1996 can be solved by

forcing one variable to the boundary or not? This can be answered by considering the

following problem.


152

Problem 4.3. (Le Trung Kien) Given a, b, c ≥ 0, ab + bc + ca = 1 (*). Prove that

1 1 1 122a b b c c a

+ + ≥ ++ + +

Proof

First of all, consider the case c = 0, we need to prove

A = 1 1 1 122a b b a

+ + ≥ ++

, where ab = 1

We leave this simple exercise for readers.

Now, we will prove ( ) ( )1, , 0, ,f a b c f a ba b

≥ + +

Note that ƒ(a, b, c) in 4.3 is quite complicated therefore it is difficult to use MV method by

considering the difference. Fortunately, ƒ(a, b, c) can be expressed in terms of x a b= + and c

as following:

2 2

2 2

( ) ( )1 1 1 1( , , )( )( )

2 2 1 2 2 11 1 ( )1 1

c a b cf a b c


c a b c c x c g ca b xc c

+ + += + + = +

+ + + + + +

+ + + + + + += + = + =+ + +

We have:2 2

2 3 2

1 1( ) 0

( 1) (2 2 1)

cx c c cg c

c c x c

− − − +′ = ≤+ + + +

Therefore, g(c) is decreasing. Note that: 1

1cx ca cb ca cb ab cx

= + ≤ + + = ⇔ ≤ .

Thus: ( ) ( )2

1 1 1( ) , , 0, ,1

xg c g x f a b c f a b

a bx xx� �≥ = + + ≥ + +� � +

The problem is then solved. Equality occurs ⇔ 0, 1a b c= = = or its permutation

Remark:

� The technique of transforming expressions of a, b, c into expression of ,a b c+ if

sophisticated. Note that a conditioned inequality of three variables is equivalent to an

unconditioned inequality of two variable. The above technique helps us to utilize the given

condition when transforming it to inequality of 2-variables. Based on this idea, we have a

“surprised” proof for Iran TST 96 inequality.


153

With the condition 1ab bc ca+ + = . Let x a b= + , we have

( ) ( ) ( ) ( )2 2

2 2 2 2 22

2 4 21 1 1 1 ( )1

x c cx g cxa b b c c a c

+ + −+ + = + =+ + + +

This is a function of two variables.

Differentiate g(c) we have ( )

2 2 3

32

4 (3 1) 3( )

1

cx c x c cg c

c

� �+ − + −� �′ = −+

Consider following cases:

Case 1: 1c ≥ , then: 1c x≥ ≥

We have 2

21 4 44x

ab bc ca cx x cx= + + ≤ + ⇔ ≥ − , therefore

2 2 3 2 3 3 2(3 1) 3 (4 ) (3 1) 3 0cx c x c c c cx c x c c c c c x x+ − + − ≥ − + − + − = + − − ≥

( ) 0g c′⇔ ≤ . So: ( ) ( )1 1( ) 0, ,g c g f a bx a b

≤ = ++

.

Case 2: 1c ≤ . Again, we have: 2

2 44 4

4x

x cx cx

−≥ − ⇔ ≥ .

Consider the second derivative of g(c):

( )( )

2 2 2 4 2

32

4 5 1 12 ( 1) 3 18 3( )

1

c x c c x c cg c

c

� �− + − + − +� �′′ =+

We will prove: ( )2 2 2 4 2( ) 0 ( ) 5 1 12 ( 1) 3 18 3 0g c h x c x c c x c c′′ ≤ ⇔ = − + − + − + ≤

Since h(x) is a quadratic function with positive coefficient, we have ( ) ( ){ }1( ) max 0 ,h x h hc

≤ .

Also ( )4 2 4 22

11(0) 3 18 3 0 ; 3 6 4 0h c c h c cc c

= − + ≤ = − − − ≤

Therefore ( ) 0 ( ) 0h x g c′′≤ ⇔ ≤ , so:

( ) ( ){ } ( ) ( ){ }241 1( ) max , ( , , ) max 0, , , , ,4

xg c g g f a b c f a b f t t cx x a b

−≤ ⇔ ≤ ++

We now only to prove the problem when there are two variables with the same value or one

variable is 0. The first case is proved above. In the second case, we assume c = 0, thus the

condition becomes ab = 1. We have

( )2 2 2

91 1 14b aa b

+ + ≥+

2 2 2( ) (8 8 15) 0ab a b a b⇔ − + + ≥

The problem is now solved completely.


154

� We will now show examples that MV method where variables are equal cannot be applied.

Applying MV method where one variable moves to the boundary is suitable.

Problem 4.4. (Jackgarfukel)

Given a, b, c ≥ 0. Prove that 54

a b c a b ca b b c c a

+ + ≤ + ++ + +

(*)

Proof

We now consider when the equality occurs. It is easy to see that if a = b = c then the equality

does not occur. We may assume that c = 0, then (*) becomes 54

a b a ba b

+ ≤ ++

(1)

Let a + b = 1. We get (1) ⇔ ( ) 25 11 04 2

b b b− + ≤ ⇔ − ≥ (ans)

Thus equality occurs for a = 3b > 0, c = 0 (and any cyclic permutation).

Without lost of generality, assume that a = max{a, b, c}.

Normalize so that 1a b c+ + = .

Let 2

a ct += and 2

a cs −= , thus a = t + s, c = t – s, b = 1 – 2t.

Then (*) ⇔ ( ) 1 2 541 1 2

t s t t sf ss t t s t

+ − −= + + ≤+ − − −

(2)

We shall prove that ƒ(s) ≤ max{ƒ(0), ƒ(t)} for s∈ [0, t]. We have:

( )( ) ( )

3 32 2

1 21 11 22 1 2 1

t s tf ss t ts t t s

+ −′ = − + −+ − + − − −

and

( )( )

( )( )

( )( )

3 5 52 2 2

3 3 1 21

1 4 1 4 1

t s tf ss t s t t s

+ −′′ = − + ++ − + − − −

. By using 1 2 0b t= − ≥ we obtain

( )( )

( )

( )

( )

( ) ( )

( )

( )5 7 7 7 72 2 2 2 2

15 15 1 2 15 1 29 18 3 33 04 1 8 1 8 1 8 1 8 1

t s t ts tf ss t s t t s s t t s

+ − −+ −′′′ = − + = + >+ − + − − − + − − −

Because ( ) 0f s′′′ > ∀s∈[0, t] then it follows from Rolle theorem that equation ƒ′(s) = 0 has at

most two solutions in [0, t]. On the other hand, it is easy to prove that ƒ′(0) ≤ 0 and ƒ′(t) ≥ 0,

therefore ƒ′(s) may change its sign at most once (0, t), moreover ƒ′(s) is as the following

possible cases: ƒ′(s) > 0, ∀s∈(0, t) or ƒ′(s) < 0, ∀s∈(0, t) or ƒ′(s) with form −0+ on (0, t). We

then obtain ƒ(s) ≤ max{ƒ(0), ƒ(t)} ∀s∈[0, t].

Now from ( ) 504

f ≤ and ( ) 54

f t ≤ we have ( ) ( ) ( ){ } 5max 0 ,4

f s f f t≤ ≤

� Remark: Similarly, we can prove the generalization of above problem:

"Let a, b, c ≥ 0 and k∉ (0, 1). We have:

( ) ( ) ( )

( )1 kkk k k

a b c C a b ca b b c c a

−+ + ≤ + ++ + +

where 21kC k k= + − ".


155

Problem 4.5. (Phan Thanh Nam) Given a, b, c ≥ 0. Prove that:

( ) ))()()((42222 accbbacbacba −−−++≥++

Solution

Define ( ) ))()()((4),,(2222 accbbacbacbacbaf −−−++−++= .

Without loss of generality we may suppose that },,min{ cbac = . If cba ≥≥ then the

inequality is trivial since 0),,( ≥cbaf , hence we only need to consider the case cab ≥≥ .

We then have:

)0,,(),,( bafcbaf − ( ) 0)3()(4)( 222222222 ≥−++−++−++= cbabacabbacba

Finally, 0)2()(4)()0,,( 22222222 ≥−+=−−+= babaababbabaf , hence the proof is completed

Equality holds if ( ) ( )0,,)13(,, ttcba −= where 0≥t (and its permutations).

� Applying MV method when one variable moves to the boundary is essential when

considering cyclic inequalities. But what’s about symmetric inequalities?

Problem 4.6. (Pham Kim Hung) Given a, b, c ≥ 0, a + b + c = 3. Prove that:

( ) ( ) ( )3 3 3 3 3 3 3 3 3 36a b c a b b c c a ab bc ca+ + + + ≤ + +

Proof

Without lost of generality, we may assume a ≥ b ≥ c. Put

( ) ( ) ( ) ( )3 3 3 3 3 3 3 3 3, , 36f a b c ab bc ca a b c a b b c c a= + + − + + + +

( ) ( ) ( ) ( )3 33 3, , 0 36f a b c a b c a b c a b c� �+ = + − + + +� �

We shall prove that: ƒ(a, b, c) ≥ ƒ(a, b+ c, 0). Indeed, we have:

( ) ( ) ( )

( ) ( ) ( ) ( )( ) ( )

3 33 3 3 3 3 3 3 3 3 3 3

36 36 36 36, , , ,0

ab bc ca a b c bc a b cf a b c f a b c

a b c a b b c c a a b c a b c

� + + = + + ≥ +�

≥ +�� + + + + ≤ + + +� ��

Thus it is enough to show the case c = 0, or equivalently

( ) ( )3 3 3 3 2 2 3 336 36ab a b a b a b a b≥ + ⇔ ≥ +

Letting t = ab we rewrite the inequality as: ( )2 3 227 9 36 4 3t t t t− ≤ ⇔ + ≥

Using AM – GM inequality, we have: 3 3 3 3

3 234 4 3 4 32 2 2 2t t t tt t+ = + + ≥ ⋅ ⋅ ⋅ =

Equality occurs when c = 0 and a + b = 3, ab = 2 ⇔ a = 2, b = 1, c = 0 or its permutations.


156

V. SMV THEOREM - INEQUALITIES WITH FOUR VARIABLES

We begin with a well-known problem:

Problem 5.1. (IMO SL, Vietnam) Let a, b, c, d ≥ 0, a + b + c + d = 1.

Prove that: 176127 27

abc bcd cda dab abcd+ + + ≤ + (1)

Proof

Since equality occurs for 14

a b c d= = = = or 1 , 03

a b c d= = = = (or any cyclic permutation),

normal evaluation is not suitable for this problem.

Letting ƒ(a, b, c, d) = abc + bcd + cda + dab − kabcd where 17627

k = . We have:

ƒ(a, b, c, d) = ab(c + d − kcd) + cd(a + b)

Hence, we hope that ƒ(a, b, c, d) ≤ ƒ(t, t, c, d) where 2

a bt += . Since 0 ≤ ab ≤ t2, we need to

have c+ d – kcd ≥ 0. Otherwise, if c + d − kcd < 0 then we have:

ƒ(a, b, c, d) = ab(c + d − kcd) + cd(a + b) ≤ cd(a + b) ≤ ( ) 3

13 27

c d a b� �+ + + =� ��

Thus, we may assume that ( ) ( ), , , , , ,2 2

a b a bf a b c d f c d+ +≤ . This means that we can use our

method without any additional assumptions for a, b. By the symmetry, if 2

c ds += then we have:

( ) ( ) ( ) ( ), , , , , , , , , , , ,f a b c d f t t c d f t t s s f t s t s≤ ≤ = ≤

( ) ( ) ( )1 1 1 1 1, , , , , , , , ,2 2 2 2 2 2 4 4 4 4 27

t s t s t s t s t s t sf t s f f+ + + + + +≤ ≤ = = � (ans)

� Remark:

a) In this proof, we split our inequality into two cases: one solved by MV method, one is

obviously true. We also make the proof more clear by supposing the existence of

( )0 0 0 0, , ,a b c d such that ( )0 0 0 01, , ,27

f a b c d > . Moreover, we also suppose that:

( ) ( ){ }1, , , max , , , ,27 2 2

a b a bf a b c d f c d+ +≤ (1)

Problem 5.2. Given 1 2, , ..., 0na a a ≥ such that 1 2 ... na a a n+ + + = . Prove that

( )2 2 2 21 2 1 2( 1) ... ...n nn a a a na a a n− + + + + ≥

It is simple when n = 2 or n = 3. We now prove the inequality when n = 4:

Given , , ,a b c d ≥ 0 such that 4a b c d+ + + = then we have:

( )2 2 2 23 4 16a b c d abcd+ + + + ≥ (1)


157

Proof

WLOG, assume a b c d≤ ≤ ≤ . Let ( )2 2 2 2( , , , ) 3 4 16f a b c d a b c d abcd= + + + + − .

Consider the difference: ( ) 23( , , , ) ( , , , )

2 2 2c d c d

f a b c d f a b ab c d+ + � − = − − �

�

Since a b c d≤ ≤ ≤ we have 2

14

a b c dab abcd

+ + +� ≤ ≤ = ��

.

Therefore: ( , , , ) ( , , , )2 2

c d c df a b c d f a b

+ +≥ . Replace 4c d a b+ = − − , we need to prove:

( ) 22 2 24

( , , , ) 3 (4 ) 16 02 2 2

c d c d a bf a b a b a b ab

� + + − −= + + + − − − ≥ ��

Let ,2

a bx y ab+= = the above inequality is equivalent to ( )22( ) (2 8 5) 3 2 0h y x x y x= − + + − ≥

If 22 8 5 0x x− + ≥ then this is true. If 22 8 5 0x x− + ≤ then because 2y x≤ ,

( )22 2 2 2( ) (2 8 5) 3 2 2( 1) ( 2 2) 0h y x x x x x x x= − + + − = − − + ≥ . Thus (1) is true

Equality occurs ⇔ 1a b c d= = = = or 40,3

a b c d= = = = or its permutations.

Remark:

Solutions of problem 5.1 and 5.2 with n = 4 are simple and beautiful. However, we cannot see

how the method used in these solutions can be applied to other inequalities with 4 or more

variables. We now present a general method:

1. SMV [Strongly Mixing Variables] Theorem 1 [SMV]

Let ( ){ }1 2, , , ... | , ,nn i i iD D x x x x x x x ns const∈ = = ≥ α ≥ α = =� �� and ( )0 , , ...,s s s s D= ∈

• Consider the mapping T: D → D as following

Given ( )1 2 0, , ... ,na a a a D a s= ∈ ≠ , we select a pair of i j≠ (depends on function ƒ below)

such that i ja a≠ , then replace ,i ja a by its average.

• :f D → � is a continuous function such that: ( ) ( )( ) ,f a f T a a D≥ ∀ ∈ .

Then: ( ) ( )0 ,f a f s a D≥ ∀ ∈

(The proof of this theorem will be presented in the next section).

Based on SMV theorem, consider problem 5.2 with 4n ≥

Let ( )2 2 2 21 2 1 2 1 2( , ,..., ) ( 1) ... ...n n nf a a a n a a a na a a n= − + + + + − and assume that 1 2 ... na a a≤ ≤ ≤ .

Consider: 22 21 2 1 3 1 2 1 3 1

1( , ,..., ) , , ,..., , ( ) ...

2 2 2 4n n

n n n na a a a n n

f a a a f a a a a a a a a− −+ + −� � − = − − � �

� �

If 1 3 12( 1)

... nn

a a an −− ≥ then 2 2

1 2 1 3 1( , ,..., ) , , ,..., ,2 2

n nn n

a a a af a a a f a a a −

+ +� ≥ ��

.


158

We also have ( ) ( ) 32

1 2 1 3 1 1 3 12 2... 2 ... ( 2) 2 ...3 3

nn

n n nn nn a a a a a a n a a an n

−−

− −− −= + + + ≥ + + + ≥ − ⋅− −

2 3

1 3 12 5

( 3)...

2( 2)

n n

nn

n na a a

n

− −

−−

−⇔ ≥−

We will prove that: 2 3

2 5 1 32 5

2( 1) ( 3)4( 1)( 2) ( 3)

2( 2)

n nn n n

n

n n nn n n n

n n

− −− − −

−

− −≥ ⇔ − − ≥ −−

Using AM-GM: ( ) ( ) ( )2( 2)

2 22 4 3 1( 2) 3 12

nn nn n nn n n

−− −− − + −− = ≥ − −

Thus, we only need to prove ( ) 11 1 1 14( 3)( 1) ( 2) 4 1 1 0

3

nn nn n n n

n n

−− −− − ≥ − ⇔ − − + ≥

−

This is true because ( ) ( )1 4 1 271 14 1 4 1 14 16

n

n

− −− ≥ − = >

Therefore 2 21 2 1 3 1( , , ..., ) , , , ... ,

2 2n n

n n

a a a af a a a f a a a −

+ +� ≥ ��

. By SMV theorem, we only need

to consider the case 1 2 3( 1) , .... na n n x a a a x= − − = = = where 01

nx

n≤ ≤

−.

The inequality becomes: ( )( ) ( ) ( )( )2 2 1 2( 1) 1 1 1 0nn n n x n x nx n n x n−� �− − − + − + − − − ≥� �

( ) ( )2 21 2 2( 1) 1 2 1 2 0n nn x nx n x n x n n−⇔ − − − − + − + − ≤

Let ( ) ( )2 21 2 2( ) ( 1) 1 2 1 2 0n ng x n x nx n x n x n n−= − − − − + − + − ≤ . We have

( ) ( )2 21 2 2( ) ( 1) ( 1) 2 1 2 1 ( 1)( 1) 2( 1)n n ng x n n x n n x n x n n x nx n− − −� �′ = − − − − − + − = − − − −� �Hence

( )g x′ has two real solutions 1x = and ( )2 2 1 1n nx

n− −= ≥ .

It is easy to see that ( ){ }( ) max (1),1

ng x g gn

≤−

On the other hand ( )(0) 01

ng gn

= =−

so ( ) 0g x ≤ .

The problem is solved completely.

Equality occurs ⇔ 1 2 ... 1na a a= = = = or 1 20, ...1n

na a a

n= = = =

−.

Through the above examples, you might see the power and fineness of SVM. Now we will consider problem 5.1 using SMV theorem.

WLOG, assume that a b c d≤ ≤ ≤ .

Consider ( ) ( ) ( )176 176, , ,27 27

f a b c d abc bcd cda dab abcd ac b d bd a c ac= + + + − = + + + −

From the given conditions, we have: ( )1 12 2

a c a b c d+ ≤ + + + = ,

therefore 1761 1 4 827a c a c

+ ≥ ≥ ≥+

� ( ) ( ), , , , , ,2 2

b d b df a b c d f a c+ +≤


159

By SMV theorem, we only need to prove with the case 1 3 ,a t b c d t= − = = =

Replace 1 3a t= − , we have 2 3 31761327 27

at t at+ ≤ + ( ) ( ) ( )21 3 4 1 11 1 0t t t⇔ − − + ≥


a b c d= = = = or 1 , 03


Through the application of SMV with two problems, one with n variable and one with 4

variables, you can see that it is simple with 4 variables while MV method with n variables is

really complicated. Therefore, SMV is a perfect choice for inequalities with 4 variables

(given that MV method is applicable), and on the other hand 4-variable inequality is the one

use SMV the most. Let reconsider the problem 2.2.1 to affirm this:

Based on the solution, we see that the condition guarantees ( )( , , , ) , , ,2 2

c d c df a b c d f a b + +≥

is that 32

ab ≤ .

And if a b c d≤ ≤ ≤ then we have 1ab ≤ . However, we also notice that ac ab≥ and 1ac ≤ .

Thus we can use SMV and then just prove the problem where three variables are equal.

In general, for all symmetric 4-variable problems, if we have ( , , , ) ( , , , )2 2

c d c df a b c d f a b

+ +≥

for a b c d≤ ≤ ≤ then we also have ( , , , ) ( , , , )2 2

b d b df a b c d f a c

+ +≥ , therefore do we only

need to prove the case where three variable are equal. This property is not always true for

inequalities with n variables, even it is true then the proof is not simple. In the following, we

thus only introduce applications of SMV in inequalities with 4 variables.

Problem 5.3. (Phan Thanh Nam) Given a, b, c, d ≥ 0 and 4a b c d+ + + = .

Prove that: ( ) ( ) ( ) ( )2 2 2 2 8abc bcd cda dab abc bcd cda dab+ + + + + + + ≤

Proof

Letting ƒ(a, b, c, d) = ( ) ( ) ( ) ( )2 2 2 2abc bcd cda dab abc bcd cda dab+ + + + + + +

Assume a ≥ b ≥ c ≥ d. Consider the difference: ( ) ( ), , , , , ,2 2

a c a cf b d f a b c d+ + −

= ( ) ( ) ( ) ( )2 2

2 2 2 222 2

a c a cb d ac b d b d� �� − ++ + + + −� ��

( ) ( )2

2 24 2 02

a b b d abcd b d−≥ + + − ≥ (since 2 2abcd b d≥ )

Therefore: ( ) ( ), , , , , ,2 2

a c a cf a b c d f b d+ +≤ .

By SMV theorem, we only need to prove with the case , 4 3a b c x d x= = = = − , 403

x≤ ≤ :

( )3 2 4 2 6 2 4 3 23 (3 3 ) 3 (4 3 ) 8 ( 1) 28 16 12 8 0x x x x x x x x x x+ − + − + ≤ ⇔ − − − − ≤


160

It is easy to prove: 4 3 228 16 12 8 0x x x− − − ≤ with 403

x≤ ≤ .

Thus, the problem is solved. Equality occurs if and only if 1a b c d= = = = .

Problem 5.4. Given , , , 0a b c d ≥ such that 1a b c d+ + + = . Prove that:

4 4 4 4 148 127 27

a b c d abcd+ + + + ≥

Proof

Assume a ≥ b ≥ c ≥ d. and let ( ) 4 4 4 4 148 1, , ,27 27

f a b c d a b c d abcd= + + + + −

Consider the difference ( ) ( ) ( ) ( )2 27 37, , , , , , 32 2 8 27

a c a cd f a b c d f b d a c ac bd a b+ + � �= − = − + − −� ��

Since ( ) ( )0 , , , , , ,2 2

a c a cac bd d f a b c d f b d+ +≥ � ≥ � ≥ .

By SMV theorem, we only need to consider the case 13

da b c −= = = . We have:

( ) ( ) ( ) ( ) ( )4 3 241 148 1 2 4 1 19 201, , , 0

27 729 27 729d d d d d df a b c d d− − − += + + − = ≥


a b c d= = = = or 1 , 03


Problem 5.5. Given a, b, c, d ≥ 0 and a + b + c + d = 4. Prove that

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 21 1 1 1 1 1 1 1a b c d a b c d+ + + + ≥ + + + +

Proof

Let ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2( , , , ) 1 1 1 1 1 1 1 1f a b c d a b c d a b c d= + + + + − + + + +

And assume that a b c d≤ ≤ ≤ . We will prove: ( ) ( ), , , , , ,2 2

a c a cf a b c d f b d+ +≥ .

Indeed, since 2a c+ ≤ we have

( ) ( ) ( ) ( )2 22

22 2 ( ) 411 1 1 02 2 16

a c aca ca c a c� � � + +++ + − + = − − ≥ ��

Using AM – GM, ( ) ( ) ( )2

1 1 12

a ca c ++ + ≤ + .

Therefore ( ) ( ), , , , , ,2 2

a c a cf a b c d f b d+ +≥

Now we only need to consider the case , 4 3a b c x c x= = = = − :

( ) ( ) ( ) ( ) ( )3 2 32, , , 4 3 1 1 4 3 1 5 3f x x x x x x x x� �− = + + − − + −� �

( )( ) ( )( )6 4 2 2 3 2

8 7 6 5 4 3 2

3 3 1 9 24 17 3 3 1 5 3

9 24 44 72 81 68 54 36 12

x x x x x x x x x

x x x x x x x x

= + + + − + − + + + −

= − + − + − + − +


161

( ) ( )

( ) ( ) ( ) ( )

26 5 4 3 2

2 2 2 24 4 2 2

9 6 23 20 18 12 12 1

3 1 2 5 2 1 10 3 2 1 0

x x x x x x x

x x x x x x x x

= − + − + − + −

� �= − + + − + + − − ≥� �

The problem is thus solved. Equality occurs if and only if 1a b c d= = = =

Problem 5.6 [Tukervic inequality] Given , , , 0a b c d ≥ . Prove that

4 4 4 2 2 2 2 2 2 2 2 2 2 2 22a b c abcd a b b c c d d a a c b d+ + + ≥ + + + + + (1)

Proof

Assume that a b c d≥ ≥ ≥ . Let ( ) ( ) ( ), , , 1 1f a b c d LS RS= −

( ) ( )4 4 4 4 2 2 2 2 2 2 2 22a b c d abcd a c b d a c b d= + + + + − − − + +

� ( ) ( ) ( ) ( )( )2 2 2 2, , , , , , 0f a b c d f ac b ac d a c a c b d− = − + − − ≥

By S.M.V, we only need to consider the case a b c t= = = .

The inequality ⇔ 4 4 3 4 2 2 4 3 3 2 23 2 3 3 3t d t d t t d d t d t d t d+ + ≥ + ⇔ + + ≥

This is true due to AM – GM.

Equality occurs ⇔ a b c d= = = or , 0a b c d= = = or its permutations

Problem 5.7 (Pham Kim Hung) Given , , , 0x y z t ≥ such that 4x y z t+ + + = .

Prove that: ( ) ( ) ( ) ( )1 3 1 3 1 3 1 3 125 131x y z t xyzt+ + + + ≤ +

Proof

Let ( ) ( ) ( ) ( ) ( ), , , 1 3 1 3 1 3 1 3 131f x y z t x y z t xyzt= + + + + −

WLOG, we assume x y z t≥ ≥ ≥ . Consider the difference:

( ) ( ) ( ) ( )( )( )2

, , , , , , 131 9 1 3 1 32 2 4

x zx z x zf x y z t f y t yt y t−+ +− = − + +

Since x y z t≥ ≥ ≥ we have 2y t+ ≤ , this leads to ( )( )9 1 3 1 3 131y t yt+ + ≥

Therefore ( ) ( ), , , , , ,2 2

x z x zf x y z t f y t+ +≤ .

By S.M.V, we only need to consider the case 1 4 3x y z a t a= = = ≥ ≥ = − :

( ) ( )( ) ( )3 31 3 1 3 4 3 125 131 4 3a a a a+ + − ≤ + − ( ) ( )( )21 3 4 50 28 0a a a⇔ − − + ≤

This is obviously true.

Equality occurs if 1x y z t= = = = or 4 , 03

x y z t= = = = or its permutations.

� Now, we would like to conclude the section on MV method for “specific” inequalities

(inequalities with 3 or 4 variables) to move on a new section with n-variable inequalities. You

will see that, this is much more difficult. However, the main methods are developed from

inequalities with 3 or 4 variables.


162

VI. MV VIA CONVEX FUNCTION

Convex function plays an important role in the theory of inequalities. We now recall some

basic definitions:

1. Definition: Function ƒ: [a, b] → � is called convex if:

( )( ) ( ) ( ) ( ) [ ] [ ]1 1 , , , , 0,1f tx ty tf x t f y x y a b t+ − ≤ + − ∀ ∈ ∀ ∈

2. Properties:

2.1. Assume that f has second derivative in interval (a, b), then ƒ is convex on [a, b] if

( ) ( )0 , ,f x x a b′′ ≥ ∀ ∈ .

2.2. If ƒ is convex on [a, b] then ƒ is continuous on [a, b]. On the other hand, if ƒ is

continuous on [a, b] then f is convex on [a, b] if ( ) ( )

2 2f x f yx y

f++� ≤ �

� , ∀ x, y ∈ (a, b).

2.3. Jensen inequality: Assume that ƒ is a convex function in [a, b]. Then we have

(i) ( ) ( ) ( )1 21 2 ...... nn

f x f x f xx x xf

n n

+ + ++ + +� ≤ �

� ∀ x1, x2,... , xn ∈ [a, b].

(ii) Let ix be n numbers in [a, b ] and iλ be n non-negative numbers with 1

1n

ii=

λ =� , we have:

( ) ( ) ( ) ( )1 1 2 2 1 1 2 2... ...n n n nf x x x f x f x f xλ + λ + + λ ≤ λ + λ + + λ

We now show you how to apply convex function, property 2.2 and Jensen inequality, in

MV method.

Problem 6.1. Let x, y, z be real numbers such that 1x y z+ + = . Prove that

2 2 29

101 1 1yx z

x y z+ + ≤

+ + + (1) (Poland 1992)

Proof

Consider ( )21

tf tt

=+

then (1) ⇔ ( ) ( ) ( ) 33

x y zf x f y f z f

+ +� + + ≤ ��

.

We now prove that f concave or –f is convex. We have: ( ) ( )( )

2

32

2 3

1

t tf tt

−′′− =+

then ( ) 0f t′′− ≥ ,

0, 3t � �∀ ∈ � �. Thus if , , 0, 3x y z � �∈ � � then done. In the remaining cases, for the simplicity,

we can assume x ≥ y ≥ z. Since x + y + z = 1 and x, y, z ∉ [0, 1] then z < 0 which yields ƒ(z) < 0.

If 12

y < then 2 2 2

91 2 02 5 101 1 1

yx zx y z

+ + < + + =+ + +

thus it is sufficient to prove that 12

y ≥


163

If 102

z≥ ≥ − together with 12

y ≥ then:

( ) ( )2 2 2 2 294 4

5 5 101 1 1 1 11 12 2

y yx xz zx y z

+ + ≤ + + = <+ + + + +

Hence, it is enough to consider 12

z < −

If 1 32

z− > ≥ − then: 2 2 2

3 7 91 12 2 10 10 101 1 1

yx zx y z

+ + ≤ + − = <+ + +

If z < −3 then: 2x ≥ x + y = 1− z ≥ 4 so x ≥ 2 and therefore:

2 2 292 1 0

5 2 101 1 1yx z

x y z+ + < + + =

+ + +(ans).

Remark: If two of x, y, z are in 0, 3� �� then using MV method, they are equal. Using convex

function in applying MV method is a good choice but it is suitable when the inequality has form

( ) ( ) ( )1 2 ... nf x f x f x+ + + ). Otherwise, we need more trick. Let us explorer more details.

Theorem: Let f : [a, b] → � be a convex function. Then we have:

( ) ( ) ( ){ } [ ]Max , , ,f x f a f b x a b≤ ∀ ∈

Proof: Since ƒ is continuous then ƒ attaints its maximum value at [ ]0 ,x a b∈ .

Consider 0 0x a x b− ≤ − � [ ]1 02 ,x x a a b= − ∈

From the definition, we have: ( ) ( ) ( )11 02 2

2a x

f a f x f f x+�

+ ≥ = ��

it follows that ( ) ( )0f a f x= . The case 0 0x b x a− ≤ − is similar.

Problem 6.2. Given 0a b c≥ ≥ ≥ . Prove that: ( ) 232 3a c a b c abc− ≥ + + − ⋅

(USA TST 2004)

Proof

Let ( ) ( )233 2f b a b c abc a c= + + − ⋅ − −

Since ( )3

22 0

3abcf b

b⋅′′ = > , we have ( ) ( ) ( ){ }max ,f b f a f c≤ . In addition:

( ) ( )23 22 3 2f a a c a c a c= + − ⋅ − − ( )3 23 4 0c a c ac= − + ⋅ − ≤ (AM – GM)

( ) ( )23 22 3 2f c a c ac a c= + − ⋅ − − ( )3 23 4 0a ac ac= − + ⋅ − ≤ (AM – GM)

The problem is thus solved. Equality occurs ⇔ a b c= =

The generalization with n variables of the above problem can be stated as following:

Problem 6.3. Given 1 2 ... 0na a a≥ ≥ ≥ ≥ . Prove that

( )2

1( 1) nn a a− − ≥ 1 2 1 2... ...nn na a a n a a a+ + + − ⋅


164

We now introduce two problems which are quite difficult to solve.

Problem 6.4. Let 0 < p < q, and [ ], , 1,2,..,ix p q i n∈ = be real numbers. Prove that:

( ) ( )22

21 2

1 2

1 1 1... ...4n

n

p qnx x x nx x x pq

−� �� + + + + + + ≤ + � � ��

Here we denote by [x] the greatest integer number lest than or equal to x

Proof

Since [ ],ix p q∈ , it is easy to see that the maximum value attains if { },ix p q∈ for every i.

We now assume that there are k numbers ix which equals to p and there are n − k numbers

which equals to q. Then the left hand side is equal to: ( ) k n kkp n k qp q

−� � �+ − +� � ��

( ) ( )22 p qk n k k n k

q p� = + − + − + ��

( ) ( ) ( )2

22 2 21 24

p qn k n k n n n k

pq− � �= + − = + − −� �

Since k is a integer number then ( )22 22n n k n− − ≤ (n is even) and ( )22 22 1n n k n− − ≤ − (n is odd)

This completes our present proof.

For every i, we denote the left hand side by ( )if x a function of ix , we now prove that:

( ) ( ) ( ){ }max ,if x f p f q≤ . And equality holds if { },ix p q∈ .

We have: ( ) Bf x Ax Cx

= + + . We can use derivation tool and easily to point out that equality

occurs for { },ix p q∈ . Here is a different approach.

Note that: ( ) ( ) ( ) ( ) ( ) ( );i i i ii i

B Bf x f p x p A f x f q x q Ax p x q

� � − = − − − = − − � ��

Since then if ( ) ( ) ( ){ }max ,if x f p f q> then { },ix p q∉ and 0, 0i i

B BA Ax p x q

− > − <

i i

B BAx p x q

� < < a contradiction p < q. Thus ( ) ( ) ( ){ }max ,if x f p f q≤ .

Let us consider the case of equality: Assume that ( ) ( ) ( ){ }max ,if x f p f q= for { },ix p q∉ .

If ( ) ( )if x f p= then i i

B BAx p x q

= > , therefore ( ) ( ) 0if x f q− < (a contradiction).

If ( ) ( )if x f q= then i i

B BAx q x p

= < , therefore ( ) ( ) 0if x f p− < (a contradiction).

Thus ( ) ( ) ( ){ }max ,if x f p f q= which is equivalent to { },ix p q∈


165

Remarks: (Generalized problem): Let [ ] [ ], , ,i ia a A b b B∈ ∈ for 0 < a ≤ A and 0 < b ≤ B.

Find the maximum value of ( ) ( )2 2 2 2

1 1

1 1

... ...

...n n

n n

a a b bT

a b a b

+ + + +=

+ +

Ones can easy see that an upper bound of T is 2

14

abABab AB

� + ��

by Polya.

It is natural to ask what the upper bound of AM-GM inequality is?

Problem 6.5. (Phan Thanh Nam) Let 0 < p < q, and n numbers [ ],ix p q∈ .

Prove that: ( )2

1 2

2 3 1

...2

nxx x p qnT nx x x pq

−� �= + + + ≤ +� ��

Proof

If follows from the above properties of function ( ) Bf x Ax Cx

= + + that for every i, by

replacing ix by p or ix by q T is increasing. When T is constant, { }1 2, ,..., ,nx x x p q∈ .

After finite replacing, we get { }1 2, ,..., ,nx x x p q∈ . If 1 2 ... nx x x q= = = = then T n= is the

minimum value. In order to obtain the maximum value, we just assume the existence of

ix p= . Without lost of generality, we can suppose that 1x p= . After replacing 2x q= we

substitute 3x p= . At the end of this process, we obtain

2p qnTq p

� = + ��

(if n is even) and 1 12

p qnTq p

� −= + + ��

(if n is odd)

or equivalently ( )2

,2

p qnT n npq−� �= + ∀

� ��

Equality occurs if 2ix q= , 2 1ix p+ = .

Comment: You can see that the idea of “mixing variables” turns up early even with classical

approach. Although classical inequalities are not powerful tools, however we can “stand on

the shoulder of the giants”. We have seen two important MV methods in previous sections

(centre and boundary). In special cases when optima is at the centre, convex functions give us

another interesting MV method which will be introduced in the next section.


166

VII. UNDEFINED MIXING VARIABLES – UMV

MV method with convex functions is useful in many problems. However, its disadvantage is

that it can not be used when functions of variable are not explicit or differentiation if too

complicated. Undefined mixing variables method is designed to deal with this issue

UMV − Undefined Mixing Variables Theorem: Given

� ( ){ }1 2, , ... | 0, 1,...,nn iD x x x x x i n⊂ = ∈ ≥ ∀ =� , D is closed and bounded. Let Λ be the set of

elements in D such that t coordinates are 0 and the rest are equal ( )0t ≥ .

• 2 mappings 1 2, :T T D D→ such that: For each element ( )1 2, , ... \na a a a D= ∈ Λ , select 2

indices i j≠ such that { }min 0, 1,...,i ta a t n= > = and { }1 2max , ,...j na a a a= , then replace

,i ja a by ( ), ,i ja aα β∈ (corresponding to 1T ) and i ja a′ ′α < < < β (corresponding to 2T ).

• :f D →� is continuous such that: ( ) ( )( ) ( )( ){ }1 1min , ,f a f T a f T a a D≥ ∀ ∈

then ( ) ( ){ }min ,y

f x f y x D∈Λ

≥ ∀ ∈

The theorem seems to be very abstract; however the essence is quite simple.

We begin to see applications of UMV with the following problem.

Problem 7.1. [Le Trung Kien] Given , , 0a b c ≥ . Prove that:

( ) ( )3 3 3 9 4 8a b c abc a b c ab bc ca+ + + + + + ≥ + + (1)

Proof

Let ( ) ( ) ( )3 3 3, , 9 4 8f a b c a b c abc a b c ab bc ca= + + + + + + − + +

First of all, we consider the difference (i.e. see if two variables are equal):

( ) ( ), , , ,2 2

b c b cf a b c f a + +− ( ) ( )2

3 3 9 84

b cb c a −= + − +

We can not conclude anything. We now try to see if a variable is zero:

( ) ( ), , , ,0f a b c f a b c− + ( )3 3 9 8b c a bc= − + − +

Since ( )3 3 9 8b c a+ − + is either non-negative or non-positive, therefore

( ) ( ) ( ){ }, , min , , ; , , 02 2

b c b cf a b c f a f a b c+ +≥ +

Using U.M.V: ( ) ( ) ( ) ( ){ }, , min , , ; , , 0 ; , 0, 0f a b c f x x x f y y f z≥

where ; ;3 2

a b c a b cx y z a b c+ + + += = = + +


167

Now: ( ) 3, 0, 0 4 0f z z z= + ≥ ; ( ) ( ) 23 2, , 12 24 12 12 1 0f x x x x x x x x= − + = − ≥

( ) ( )23 2, , 0 2 8 8 2 2 0f y y y y y y y= − + = − ≥ � ( ), , 0f a b c ≥ .

Equality occurs ⇔ ( ) ( ) ( ) ( ), , 0, 0, 0 ; 1,1,1 ; 2, 2,0a b c = .

Comment: You can see the power of UMN variables through above example. When normal

MV techniques does not work, combination of these techniques actually solves the problem.

Problem 7.2 [Le Trung Kien] Given , , , 0a b c d ≥ such that 4a b c d+ + + = .

Prove that: ( )2 4a b c d abc bcd cda dab+ + + ≥ + + + +

Proof

Let ( )2 4a b c d abc bcd cda dab+ + + − − − − − . Consider the difference:

� ( ) ( ), , , , , ,2 2

a b a bf a b c d f c d+ +− ( )( ) ( ) ( )2

2 24

a ba b a b c d ab� += + − + + + − ��

( )

( ) ( )( )( ) ( )

2 2

28

4 42

a b a bc d c d Xa b a b a b

− −� �= + − = + −� �+ + + +� �

� ( ) ( ), , , , 0, ,f a b c d f a b c d− +

( ) ( )( )

( )42 a b a b c d ab c d ab Y c dab a b a b

� �= + − + − + = − − = − −� �+ + +� �

It is easy to see that: ( ) ( ) ( )( )284

2ab a b a b a b a b a b≥

+ + + + + + +

or equivalently Y X≥ thus c d Y

c d X

+ ≤�� + ≥�

� ( ) ( ) ( ){ }, , min , , , ; , 0, ,2 2

a b a bf a b c f c d f a b c d+ +≥ +

By U.M.V theorem we have

( ) ( ) ( ) ( ) ( ){ }4 4 4, , , min 1,1,1,1 ; , , , 0 ; 2, 2, 0, 0 , 4, 0, 0,03 3 3

f a b c d f f f f≥ 0=

The problem is then solved. Equality occurs ⇔ ( ) ( ) ( ), , , 1,1,1,1 ; 4, 0, 0, 0a b c d = .

Now we turn back to Problem 5.2

Problem 7.3. Given 1 2, , ..., 0na a a ≥ such that 1 2 ... na a a n+ + + = . Prove that

( )2 2 2 21 2 1 2( 1) ... ...n nn a a a na a a n− + + + + ≥


168

Proof

Let ( ) ( ) ( )2 2 21 2 1 2 1 2, , , 1n n nf a a a n a a a na a a= − + + + +� � � . We have

( ) ( ) ( )( )21 21 2 1 2

1 2 3 3 42 1, , , , , , ,

2 2 4n n n

n a aa a a a nf a a a f a a a a an

−+ +� −− = − ��

� � �

( ) ( ) ( )( )1 2 1 2 3 1 2 3 42 1, , , 0, , , ,n n n

nf a a a f a a a a na a a a an−− + = − −� � �

� ( ) ( )1 2 1 21 2 3 1 2 3, , , min , , , , , 0, , , ,

2 2n n na a a a

f a a a f a a f a a a a� �+ +� ≥ +� � �

� � ��

By UMV, we only need to prove when there is at least one number in 1 2, , , na a a� is 0 and the

rest are 1. These cases are simple enough to prove.

Remark: Examples of using UMV in inequalities with 3 variables, 4 variables and n

variables, we would like to state that UMV is independent of the number of variables. This is

because UMV uses arbitrary pair of variables. This feature of UMV helps us to solve many

complicated problems.

Problem 7.4 (Pham Kim Hung) Given 1 2, , , na a a� ≥ 0 such that 1 2 ... na a a n+ + + = .

Find the minimal value of 2 2 21 2 1 2

1 2

1 1 1... ......n nn

S a a a a a aa a a

� = + + + + + + + ��

�

Solution

Let ( ) 2 2 21 2 1 2 1 2

1 2

1 1 1, , .... .... ... ......n n nn

f a a a a a a a a aa a a

� = + + + + + + + ��

Consider the differences

� ( ) ( )1 2 1 2 3, ,.... 0, , ,....,n nf a a a f a a a a− + 1 2 3 43 4

1 1 12 ..... ......nn

a a a a aa a a

� �� = − − + + + �� (*)

� ( ) 1 2 1 21 2 3, ,.... , ,. ,....,

2 2n na a a a

f a a a f a a+ +� − �

�

( )2

1 23 4

3 4

1 1 12 ..... ......4 n

n

a aa a a

a a a− � �� = − + + + ��

(**)

� ( ) ( ) 1 2 1 21 2 1 2 3 3, ,.... min 0, , ,...., , , ,. ,....,

2 2n n na a a a

f a a a f a a a a f a a� �+ +� ≥ + ��

� � �

Thus, by UMV we have: ( ) ( )1 21,

, , .... Min , , ..., , 0, 0..., 0nk n

n n nf a a a fk k k=

≥

Therefore, the minimum value is ( ) 12 2min 2 , ,

1 1 2

nn n nnn n n

−� �+� �− − −� �


169

VIII. MV WITH MEAN VALUE

Here is another technique in applying convex function by using MV method.

Theorem 1: Let ƒ : [a, b] → � be a convex function. Then:

( ) ( ) ( ) ( ) [ ], ,f a f b f x f a b x x a b+ ≥ + + − ∀ ∈

Proof

Since [ ],x a b∈ then ( )1x ta t b= + − for [ ]0,1t ∈ . Then: ( )1a b x t a tb+ − = − +

Using the definition of convex function, we have:

( ) ( ) ( )[ ] ( )[ ]1 1f x f a b x f ta t b f t a tb+ + − = + − + − +

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )1 1tf a t f b t f a tf b f a f b� � � �≤ + − + − + = +� � � �

Theorem 2: (Jensen inequality) Let ƒ : [a,b] → � be a convex function. Then

∀ 1 2, ,..., nx x x ∈[a,b] , we have: ( ) ( ) ( ) ( )1 21 2

...... ( )n

n

x x xf x f x f x nf nf T

n+ + +

+ + + ≥ =

Proof

� Step 1: If 1 21 2

...... n

n

x x xx x x T

n+ + +

= = = = = (*) then the inequality holds.

� Step 2: If (*) does not hold, without lost of generality, we can assume that 1 2x T x> > .

By replacing ( )1 2, , ... nx x x by ( )1 2, , ..., nT x x T x+ − function ƒ is increasing. Moreover, after

processing Step 2, the number of variables which is equal to T is increasing, therefore, (*)

holds after at most ( )1n − steps.

Comments: In n-variable inequalities (n � 4), mixing variables to the arithmetic mean value is

better than mixing variables to the center. Indeed, let us consider more examples.

Problem 8.1. Let 1 2, , ... na a a > 0 and 1 2... 1na a a = . Prove that:

For k = 4(n − 1) we always have: 1 2 1 2

1 1 1......n n

k kna a a a a a n

+ + + + ≥ ++ + +

(1)

Proof

For n = 1, n = 2 the assertion is trivial, we now need to prove under the case n ≥ 3.

Proposition 1: Let ( )1 21 2 1 2

1 1 1, ,... ......n

n n

kf a a aa a a a a a

= + + + + + + + .Then we have:

(i) If 1 2a x a≤ ≤ and 1 2 1a a ≤ then ( ) 1 21 2 3, , ... , , , ...,n n

a af a a a f x a a

x�

≥ ��


170

(ii) If ( ) ( )1 2 1 2 1 21 3

1 1 1 0n n

i ii i

a a ka a a a a a= =

� �� − − − + + ≥� � �

� �� then ( ) ( )1 2 1 2 3, ,... 1, , ,...,n nf a a a f a a a a≥

(iii) If 1 2 3, 1a a a≥ ≥ then ( ) ( ) ( ){ }1 2 1 2 3 2 1 3 1, , ... min 1, , ,..., , 1, , , , ...,n n i nf a a a f a a a a f a a a a a a≥

Proof:

(i) We see that

( ) 1 21 2 3

1 2 1 2 1 2 1 2

1 1 1, ,... , , ,...,n n

a a x k kf a a a f x a ax a a x a a A a a a a

A xx

� − = + − − + − � + +� + +

( ) ( ) ( ) ( )

( ) ( )1 2

1 2

1 2 1 2 1 2

1 2 1 2

a ax

a ax

x a x a A a a A x ka a

xa a A a a A x

� �− − + + + + −� �=+ + + +

where 3

n

ii

A a=

=�

According to AM – GM: ( ) ( )21 21 2 1 24 1

a aA a a A x n n k ka a

x�

+ + + + ≥ ≥ − = ≥ ��

� (ans).

(ii) From the above inequalities, we put x = 1 then: ( ) ( )1 2 1 2 3, , ... 1, , , ...,n nf a a a f a a a a− =

( ) ( ) ( ) ( )

( ) ( )1 2 1 2 1 2 1 2

1 2 1 2 1 2

1 1 1

1

a a ka a A a a A a a

a a A a a A a a

� �− − − + + + +� �=+ + + +

which yields the assertion.

(iii) We consider two following cases: ( ) ( )1 2 1 2 1 21 3

1 1 1 0n n

i ii i

a a ka a a a a a= =

� �� − − − + + ≥� � �

� ��

Case 1: If 1 2 1 21 3

1n n

i ii i

ka a a a a a= =

� ≥ + + �

� � � then from (ii) we get ( ) ( )1 2 1 2 3, ,... 1, , ,...,n nf a a a f a a a a≥ .

Case 2: If 1 2 1 21 3

1n n

i ii i

ka a a a a a= =

� ≤ + + �

� � � since 3 21a a≤ ≤ then we have:

1 3 1 3 1 2 1 21,3 1 1 3

1 3 1 3 1 2 1 2

1 1 1 11 1

n n n

i i i ii i i i

a a a a a a a a a a a a

a a a a a a a a≠ = = =

+ + − − + − − + + += + ≥ + =

� � � �

� 1 3 1 31 1,3

1n

i ii i

ka a a a a a= ≠

� ≤ + + � �

� � � . Again by using (ii) we have: ( ) ( )1 2 2 1, ,... 1, ,..., ,...,n i nf a a a f a a a a≥ .

Remark: Proposition 1 helps us to reduce the number of variables to one.


171

Proposition 2: Inequality (1) with variables 1 2, , ... na a a >0 and 1 2 ... 1na a a = can be deduced to

the case where there are n − 1 variables which are equal and ≤ 1.

Proof:

� Step 1: Reduce to the case where there are n − 1 variables which are less than or equal to 1.

Assume that at least two variables are greater than 1, without lost of generality, we may

assume 1 2,a a . Using Proposition 1 (iii) we can replace ( )1 2, , ... na a a by another such that ƒ is

unchanged, and moreover, the number of variables which are equal to 1 is increasing at least 1. Hence, we obtain our conclusion after at most (n − 1) steps.

� Step 2: Deduce these variables are equal.

Assume that there are 1 2 1... 1na a a −≤ ≤ ≤ ≤ such that its geometric mean value equals to x.

If they are not equal, then 1 1na x a −< < . Using Proposition 1 (i) we can replace ( )1 2 1, , ... ,n na a a a−

by 1 12, , ..., ,n

n

a ax a a

x−�

��

. Then ƒ is not decreasing and then number of variables which are

equal to x is increasing at least 1. On the other hand, since 1 1 1 1na a ax x

− ≤ ≤ then 1a x≤ , this

means that we can apply this step again and again.

Now we prove the one-variable problem: ( )1

1, , ..., , 1,1,...,1n

f x x x fx −

� ≥ ��

for x ≤ 1

Letting ( )( )

11

1

11, ,..., ,11

nn

n

n kg x f x x x xxx n x

x

−−

−

−� = = + + �� − +

for x∈ (0, 1]

After some calculations, we get:

( ) ( )( )

( ) ( )( )

22

2 2 2

1

111 11 11 11 111

n nnn

n

n

nk nx n xn xg x n x n

x x n xn xx

−

−

−� �− −� � � �− −� �− −′ = − + − − = − � �− +� �� − +

� ��

Since k = 4(n − 1) then g’(x) ≤ 0 for x∈ (0, 1], it follows that g(x) ≥ g(1) which completes our proof.

Problem 8.2. Let 1 2, , ... na a a be n real numbers such that 1 2. ... 1na a a = . Prove that:

( ) ( ) ( ) ( )2 22 2 21 2 1 22 2

21 1 ... 1 ...n n

n nna a a a a a

n−

−+ + + ≤ + + +

Proof

The cases n = 1, n = 2 are trivial. We now consider the case n ≥ 3.

It is sufficient to prove ( )1 2, , ... 0nf a a a ≥ or prove ( )1 2, , ... 0ng a a a ≥ ,

where ( ) ( ) ( ) ( ) ( )2 2 2 2 21 2 1 2 1 2, , ... ... 1 1 ... 1n

n n nf a a a k a a a a a a−= + + + − + + +

( ) ( ) ( ) ( ) ( ) ( )2 2 21 2 1 2 1 2, , ... ln 2 2 ln ... ln 1 ln 1 ... ln 1n n ng a a a k n a a a a a a= + − + + + − + − + − − +


172

Proposition 1:

(i) If 1 2 31 ,a a a≥ ≥ then: ( ) ( ) ( ){ }1 2 1 2 3 1 2 3, ,... min 1, , ,..., , ,1, ,...,n n nf a a a f a a a a f a a a a≥

(ii) If { }1 1max n

i ia a

== and 1 2 1a x a≥ ≥ ≥ then: ( ) 1 2

1 2 3, , ... , , , ...,n n

a ag a a a g x a a

x�

≥ ��

Proof

(i) Consider ( ) ( )1 2 1 2 3, , ... 1, , , ...,n nf a a a f a a a a− =

( ) ( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 2 21 2 1 2 32 1 1 1 1 ... 1n n

nks ku a a a a a a− − � �= − + + − + + + +� �

(where 1 2 1 2... , 1 ...n ns a a a u a a a= + + + = + + + )

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )

2 3 2 4 2 3 2 2 2 21 2 1 2 1 2 3

2 3 2 4 2 3 2 21 2 1 2 3

1 ... 1 1 1 ... 1

1 1 ... 1 1 1 ... 1

n n nn

n n nn

k a a a a s s u u a a a a

a a k s s u u a a a a

− − −

− − −

= + − − + + + + − − + +

� �= − − − + + + − + + + +� �

Similarly, we have: ( ) ( )1 2 1 2 3, , ... ,1, ,...,n nf a a a f a a a a− =

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 2 4 2 3 2 2 22 3 2 3 1 41 1 ... 1 1 1 1 ... 1n n n

na a k s s v v a a a a a− − −� �= − − − + + + − + + + + +� �

(where 1 2 3 41 ... nv a a a a a= + + + + + )

From the above equalities, we see that:

� If ( ) ( ) ( ) ( ) ( )2 3 2 4 2 3 2 21 2 3... 1 1 1 ... 1 0n n n

nk s s u u a a a a− − −+ + + − + + + + ≥ (2)

then ( ) ( )1 2 1 2 3, , ... 1, , ,...,n nf a a a f a a a a≥ .

� If ( ) ( ) ( ) ( ) ( ) ( )2 3 2 4 2 3 2 2 22 3 1 4... 1 1 1 1 ... 1 0n n n

nk s s v v a a a a a− − −+ + + − + + + + + ≤ (3)

then ( ) ( )1 2 1 2 3, , ... ,1, ,...,n nf a a a f a a a a≥ .

Therefore, it is necessary to prove either (2) or (3).

For example, we assume that (2) not hold, we shall prove that (3) hold.

It is enough to show that u ≥ v and ( ) ( ) ( ) ( )2 21 3 3 11 1 1 1a a a a+ + ≤ + +

This is followed form some simple calculation

( ) ( )3 1 2 1 2 3 2 1 31 0u v a a a a a a a a a− = + − − = − − ≥

( ) ( ) ( ) ( ) ( ) ( )2 21 3 3 1 3 1 1 3 1 31 1 1 1 1 0a a a a a a a a a a+ + − + + = − + + − ≤

This proves Proposition (i).

(ii) Because of the presence of function ln we must use derivation tool instead of applying the

above method. Consider: ( ) ( )

( ) ( ) ( )

21 3

22 2 2 22 2

1 321

ln 2 1 ln ...

ln 1 ln 1 ln 1 ... ln 1 ,1

n

n

ag t k n ta a a

t

a at a a a t

at

� = + − + + + + ��

� �� − + − + − + − + ∈ � � �

� � ��


173

We have: ( )( )

( )

22 2 2

1 12 3

22 2 2 2

1 3 1 2

22 1 2

... 1 1n

a an a ta

t tg ta ata a a t at t

� − − − �

� ′ = −� + + + + + + ��

=

( )

21

21 2 2

2 2 2 21 3 1 2

12... 1 1n

ataa n ta

at ata a a t at t

� �+� �� −= − −� � �� + + + + + + ��

Since 2

1

,1a

ta

� �∈ � ��

then 21 2

0a

at

− ≥ . Therefore, denote by T the last factor, it is enough to

show that T ≥ 0 in order to show that g is increasing (in 2

1

,1aa

� ��

)

Denote ( )2

2 2 2 21 12

1 1 ,a a

c t a d tatt

� = + + = + �

�

We have: ( ) ( )2 232

3

10 1 ...... n

n

n dT n c d d a ad a a c

−≥ ⇔ ≥ ⇔ − ≥ + + ++ + +

Since c ≥ d (Cauchy-Schwarz inequality) then it suffices to prove that ( )32 ... nn c a a− ≥ + +

And it is true since { }1 2 1 3max ,..., nc a a a a a> ≥ ≥

Choose { }1 20

1

1max ,a a

t xx a

= ⋅ , we get { } { }2 1 2 2 1 20 0 1

1 0

,1 , max , , min ,a a a a a a

t t a x xa x t x

� �∈ = =� ��

Since g is increasing in 2

1

,1aa

� ��

then g(1) ≥ g(t0) (ans).

This completes the proof of Proposition (ii).

Turn back to our problem. We say that ( )1 2, , ... na a a is substituted by ( )1 2, , ... nb b b if

( ) ( )1 2 1 2, , ... , , ...n nf a a a f b b b≥ or ( ) ( )1 2 1 2, , ... , , ...n ng a a a g b b b≥ .

Proposition 2: We always deduce to the case where there are (n −1) variables which are

equal and ≥ 1.

Proof

� Step 1: Deduce to the case where there are n − 1 variables which are equal and ≥ 1.

Assume that there exist 2 3, 1a a < . It is easy to see that at least one of which is greater than 1,

without lost of generality, we can assume that it is 1a . By using Proposition 1 (i), we can

substitute ( )1 2, , ... na a a by ( )1 2 31, , ,... na a a a or by ( )1 2 3,1, ,... na a a a . We observe that the

number of variables which are equal to 1 is increasing, thus, we cannot substitute over n −1

times.


174

� Step 2: We now claim that we can substitute n −1 variables which are greater than or equal

to 1 by their GM. Indeed, assume that 1 2 1... 1n na a a a−≥ ≥ ≥ ≥ ≥ and put 11 2 1... 1n

nx a a a−−= ≥ . If

there exists at least one variable which is different from the first n−1 variables then

1 1na x a −> > . By using Proposition (ii), we can substitute ( )1 2 1, , ... ,n na a a a− by

1 23, , ..., , n

a ax a a

x� ��

. Since 1 11 1n

n

a aa

x−

−≥ ≥ (because 1a is the greatest number in { } 1

1

ni i

a −=

, this

means that 1a x> ) then our substitute is valid. After at most n − 1 times, the first n − 1

variable are equal to x. We now turn to the one-variable problem.

Consider the following function ( )1

1, ,..., ,n

h x g x x xx −

� = ��

( ) ( ) ( ) ( )21 2 2

1 1ln 2 1 ln 1 1 ln 1 ln 1n n

k n n x n xx x− −

� � � = + − − + − − + − + �� for x ≥ 1.

We have: ( ) ( )( )

( )( )

2 1

2

1 2 2

2 1112 12 1

1 111 1

n n

n n

nnnn xx xh x n

xn xx x

−

− −

−−− − −−′ = − − −

+− + +

( ) ( ) ( )( )

( ) ( ) ( )( ) ( ) ( )

2 2

2 2 2 2 2 2

1 1 1 12 1 2 1 111 1 1 1 1 1 1 1

n n n

n n n n

n x n xn nx xx xn x x x n x x x− −

� � � �− − − −− − −= − + = −� � � �− + + + − + + +� � � �

Since x ≥ 1 then h′(x) ≥ 0 provided: ( ) ( ) ( )2 2 2

1 11 1 1 1

n

n nn x

n x x x −− +≥

− + + +

The above inequality is easy, for example: ( ) ( ) ( )2 2 2

1 111 1 1 1 1

n

n n nn x

n x x x x −− +≥ ≥

− + + + +

Thus, for x ≥ 1 we get h′(x) ≥ 0 or equivalently h(x) is increasing,

it follows that h(x) ≥ h(1) = 0 (ans).

Equality occurs if 1 2 ... 1na a a= = = = for n ≥ 3.

Remarks:

� The above problem is an open problem by Pham Kim Hung and our proof is the first presence of it.

� By considering simultaneously functions ƒ, g, it is able to expand MV method.


175

IX. INEQUALITY GENERAL INDUCTION (IGI)

One of the mixing variable technique ((n – 1) equal variable), which was introduced by Pham Kim Hung in his book “Secrets in Inequalities” we should mention is the IGI. We may think about the ideas of this technique as follows. Suppose that we want to prove an inequality in n

variables naaa ,...,, 21 with the constraint 1...21 =naaa , where equality holds when the

variables are all equal. Because of the condition of the variables, we cannot apply the

induction method directly. However, if we modify the constraint to 1...21 ≥naaa then we may

assume that 1 2min{ , ,..., }n na a a a= to obtain 1... 121 ≥−naaa and we may then use the

hypothesis for these n-1 variables to reduce the problem to the case where n-1 variables are all equal. Following are some typical examples for this technique.

Problem (Pham Kim Hung). Given 1 2, ,..., 0na a a > and 1 2 ... 1na a a = .

Prove that: 1 2

1 1 1... min 1,

(1 ) (1 ) (1 ) 2k k k kn

na a a

� �+ + + ≥ � �+ + + � �

, ∀ 0k >

Solution

� We prove a more general result:

If 0k > and 1 2, ,..., na a a are positive numbers whose product 1≥ then:

1 2 1 2

1 1 1... min 1 ,

(1 ) (1 ) (1 ) (1 ... )k k k knn n

na a a a a a

� �� + + + ≥ � �+ + + +� �� (1)

� Induction on n .

If 1n = then (1) is obviously true! Suppose (1) holds true to n ( 1)n ≥ , we show that it is also

true for 1n + .

Let 1 2 1, ,..., na a a + be positive real numbers whose product is 1ns + where 1s ≥ ,

we need to prove: 1 2 1

1 1 1 1... min 1,

(1 ) (1 ) (1 ) (1 )k k k kn

na a a s+

+� �+ + + ≥ � �+ + + +� � (2)

We now fix s. First, let us simplify the problem.

� Clause: Let 0ln( 1)ln(1 )

nk

s+=+

(which means 0

11

(1 ) k

ns+ =

+)

Suppose (2) holds true for 0k k= , then so is every 0k > . Moreover, we only have 0k n≥ in

the case 0 1n k s= = = and then (2) is also true.

Proof: The fact that (2) holds true for 0k k= means that 0 0

1

1

1 11

(1 ) (1 )

n

k ki

na s

+

=

+≥ =+ +�

If 0k k> then:

00

00

11

11

111(1 )(1 ) (1 ) 1

1 1 1 (1 )

kknn kk

kkkii i i

k

nsa a

n n n s

++

==

� � +� � � � � �++ + � � � �≥ ≥ = � �+ + +� � +

��


176

If 0k k< then: 0

1 1

1 1

1 11

(1 ) (1 )

n n

k ki ii ia a

+ +

= =≥ ≥

+ +� �

Moreover, if 0k n≥ then 001 (1 ) 1 1kn s k s n+ = + ≥ + ≥ + (since 0 1k n≥ ≥ and 1s ≥ )

Which implies 0 1n k s= = = , then (2) is obviously true.

� Now we return to our problem. The preceding claim allows us to prove (2) for the only case

0k k= , and 0k n< . From now we will consider k to be 0k , which means that ln( 1)ln(1 )

nk

s+=+

.

Assume that 1 2 1... na a a +≥ ≥ ≥ . Then 1 2, ,..., na a a are positive numbers whose product 1≥ ,

hence by induction hypothesis

1 2

1 1 1... min 1,

(1 ) (1 ) (1 ) (1 )k k k kn

na a a a

� �+ + + ≥ � �+ + + +� � where 1 2...n

na a a a s= ≥

Since 1

1

n

n n

sa

a

+

+ = , (2) is true if we can show that 1

1 1min 1, =1

(1 ) (1 )1

k k kn

n

n na ss

a

+

+� �+ ≥ � �+ +� � �+ �

�

(3)

� Consider the LHS of (3) as a function in a and we denote it by ƒ(a). In order to prove (3),

we will investigate the behavior of ƒ where a +∈� . We have:

1

1

1 11( )

(1 )1

n

n

k kn

n

sknkn af aa s

a

+

+

+ ++

−′ = ++ � + �

�

[ ]1

( ) 0 ( ) ( 1) ln( ) ln( ) ( 1) ln(1 ) ln 1 0n

nsf s g a n s a k aa

+� �� ′ ≥ ⇔ = + − + + + − + ≥ ��

Note that: 1 1 1

1

( ) ( 1) ( 1) ( 1)( )

(1 )( )

n n n n

n n

n k a n a n ks a kn sg a

a a a s

+ + +

+− − − + + + + −′ =

+ +

( )1 1 1( ) 0 ( ) ( ) ( 1) 1 ( 1) 0n n n ng a h a n k a n a n ks a kn s+ + +′ ≥ ⇔ = − − − + + + + − ≥

We have: 1 1( ) ( 1) ( ) n n nh a n n k a na ks− +� �′ = + − − − +� �

Since k n< , ( )h a′ changes its sign from the positive sign to the negative one on +� . Moreover

h(0) > 0, hence h(a) also changes its sign from the positive sign to the negative one on +� .

Since ( ) 0g a′ = has the same sign with h(a) , the equation ( ) 0g a = has at most 2 roots on +� .

It follows that ( ) 0f a′ = has at most 2 roots on +� . Moreover, the facts that ( ) 0f s′ = and

0lim ( ) 1 ( ) lim ( )

aaf a n f s f a

+ →+∞→= ≥ = = implies that the other root of ( ) 0f a′ = on +

� is 0 ( , )a s∈ +∞ .

Therefore ƒ(a) increases on 0( , )s a and decreases on the intervals ),0( s and ),( 0 ∞a .

It follows that ( ) ( ) lim ( ) 1x

f a f s f x→+∞

≥ = = , a +∀ ∈� . The proof of the problem is completed.


177

X. ENTIRELY MIXING VARIABLES – EMV

All the MV techniques we have seen so far have one thing in common, that is if

( , , ) ( , , )f a b c f a b c′ ′ ′≥ then the tuple ( , , )a b c and ( , , )a b c′ ′ ′ must have the same characteristic,

for example sum or product, etc.

EMV technique also has unchanged quantity after “mixing variables”. However, the

difference with other methods is that the quantities used are the difference of variables, i.e.

, ,a b b c c a− − − ”.

The idea of EMV is that when we increase or decrease variables with the same value then the

inequality becomes unchanged or weakened. Thus we only need to consider the case when

one variable is on the boundary.

Not that if we use EMV for inequalities with n variables, when force one variable to the

boundary, the inequality become new inequality with n-1 variable, if this has more than 3

variables then it is still complicated. Thus, we will introduce only the application of EMV for

inequalities with 3 variables

1. EMV with the boundary at 0:

There are many inequailities where variable has boundary at 0, especially those with three

variables. In IV, we consider a technique to force the variables the the boundary at 0. Now

EMV gives us another approach. We will begin with a famous inequality

Problem 10.1 [Dao Hai Long] Given , ,a b c ∈� . Prove that

2 2 22 2 2

1 1 1 9( )

2( ) ( ) ( )a b c

b c a c b a� �+ + + + ≥� �− − −� �

Proof

Since ( ) ( ), , , ,LS a b c LS a b c≤ we only need to consider the case when 0, 0a b c≥ ≥ ≤

We have: 2 2 2 2

2 2 2 2 2 2( ) ( ) ( ) ( )( ) ( ) ( )

3 3a b c a b b c c a

a b c a b b c c a+ + − + − + −+ + = + ≥ − + − + −

Thus, it suffices to prove that

( )2 2 22 2 2

1 1 1 27( ) ( ) ( )

2( ) ( ) ( )a b b c c a

b c a c b a� �− + − + − + + ≥� �− − −� �

(1)

Let LHS (1) = ƒ(a, b, c). Notice that the above inequality contains only characteristic quantities

of EMV, therefore ( , , ) ( , , ),f a b c f a x b x c x x= − − − ∀ ∈� . We now assume a variable is 0, let

assume b, i.e. x b= the tuple ( ), 0,a b c b− − satisfies 0, 0a b c b− ≥ − ≤ .


178

Now, we need to prove: ( )2 2 22 2 2

1 1 1 27( )

2( )a c c a

c a c a� �− + + + + ≥� �−� �

(2)

We have: 2

2 22 2 2

( ) 1 1 8,

2 ( )a c

c ac a c a

−+ ≥ + ≥−

.

Thus: LHS (2) ( ) ( )2

2

(1 8) 27112 2( )

a ca c

+≥ + − =−

. The problem is then solved.

Equality occurs if , 0,a t b= = c t= − for *t ∈� or its permutations.

Remark: EMV technique is used uniquely in this problem, in which the expression is

unchanged. Let come back to Schur inequality again.

Problem 10.2 (Schur inequality) Given a, b, c ≥ 0. Prove that:

( ) ( ) ( )3 3 3 2 2 23a b c abc a b c b c a c a b+ + + ≥ + + + + + (1)

Proof

(1) ⇔ ( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 2, , 0f a b c b c a b c a c b a c a b c b a= + − − + + − − + + − − ≥ (2)

We will prove ( , , ) ( , , ),f a b c f a x b x c x≥ − − − [0, min{ , , }]x a b c∀ ∈ .

This is obviously true because b c a b c a x b x c x a x+ − ≥ + − − = − + − − + and similar results.

By EMV, we only need to consider the case when one variable is 0, which is true.

Remark: The solution of above problem use EMV under SOS form:

( ) ( ) ( ) ( ) ( ) ( )2 2 2, , , , , , 0a b cS a b c b c S a b c c a S a b c a b− + − + − ≥

This is efficient technique because it eliminates quadratic terms 2( ) ,a b− 2 2( ) , ( )b c c a− − .

Thus we only need to prove ( ) ( ), , , , , { , , }g gS a b c S a x b x c x g a b c≥ − − − ∈

However, it is not always that simple:

Problem 10.3. Given a, b, c ≥ 0. Prove that ( )91 1 1 1 1 14a b c b c c a a ba b c

+ + + ≥ + ++ + ++ + (1)

Proof

The inequality (1) ⇔ ( ) ( ) ( )1 1 1 4 3a b ca b ca b c b c c a a b

+ + + + ≥ + + ++ + +

⇔ ( ) ( )2 221 20 ( )( ) 0

( )( )b c a b c a ab ac bc b c

bc a b a c� �− − ≥ ⇔ + + + − − ≥� �+ +� �

� �

Now, you can see that the technique used in previous problem is not applicable in this

problem. Thus we need to find more suitable transformation. Note that with EMV technique,

the more , ,a b b c c a− − − occur, the more efficient the method is.


179

Thus we continue the transformation as following ( )22( )( ) 0a b c a ab ac bc b c+ + + − − ≥�

( ) ( )2 232 ( ) ( )( )( ) 0a b c b c a b c a b a c b c⇔ + − − + − − − ≥� �

( ) ( ) ( ) ( )2 2 2 232 ( ) 0a b c b c b c c a a b⇔ + − + − − − ≥�

This is obviously true. Equality occurs ⇔ a b c= = or , 0a b c= = and its permutations.

Comments: Transformation to expressions of ( ) ( ) ( )2 2 2b c c a a b− − − is nice and efficient.

You can see the tightness of that expression in estimation that closes to 0. It is much clearer

when we come back to Iran inequality

Problem 10.4. Given , , 0a b c ≥ . Prove that ( )( ) ( ) ( )2 2 2

91 1 14

ab bc caa b b c c a

� + + + + ≥ �+ + +�

(1)

(Iran TST 1996)

This problem is the most well studied and there are many solutions have been developed for

it. It is “surprised” to know that this problem can also be solved by EMV.

First of all, we carry out the following transformation

( )( ) ( ) ( )

( )

2

2 24 49

4 4

b a b c a c b cbc ca abb c b c

− + − + −+ + − =+ +

� �

( )( ) ( )

( )( )

2

2 2 214

b cc bb cc a b a b c

−� �= − − +� �+ + +� ��

( ) ( ) ( )

( )2 2

2 2 21

4bc a b c

a b a c b c

� �−= + −� �+ + +� ��

( )( )( )( ) ( ) ( )

( ) ( )( ) ( )

( )( ) ( )

( ) ( ) ( )( ) ( ) ( )

22 2

2 2 2 2 2

2 2 2 2

2 2 2 2 2

3 34

4

a b a c a bc ab ac bc b cb cb c c a a b c a a b

bc b c b c c a a bc a a b b c c a a b

− − + + + −= − ++ + + + +

− − − −= −+ + + + +

� �

�

Thus (1) ⇔ ( )

( ) ( )( ) ( ) ( )( ) ( ) ( )

2 2 2 2

2 2 2 2 24bc b c b c c a a b

c a a b b c c a a b

− − − −≥+ + + + +

�

( ) ( ) ( ) ( )2 2 2 22 24 bc b c b c c a a b⇔ − ≥ − − −�

Therefore, we only need to consider the case when one variable is 0, which is already proved

in previous section. Moreover, we can tighten the inequality as following:


180

Problem 10.5 [Le Trung Kien] Given , , 0a b c ≥ .Prove that

( )( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )

2 2 2

2 2 2 2 2 21591 1 1

4 4b c c a a bab bc ca

b c c a a b b c c a a b

− − −� �+ + + + ≥ +� �+ + + + + +� �

Again, we only need to prove when c = 0:

( )

( )( )

24

2 2 2 21591 1 1 ( ) 0

4 4a bab a b

a b a b a b−� �+ + ≥ + ⇔ − ≥� �+ +� �

Equality occurs if and only if a b c= = or , 0a b c= = or its permutations.

Notes: The above presentation is to illustrate the idea of EMV. You can also see that the

inequality is directly equivalent to

( ) ( ) ( ) ( ) ( ) ( )2 22 2 2 2 2 2 2 24 0ab a b a c b c a b ca c a bc b c� + − − − − + − + − ≥�

This is true because 0, 0a a c b b c≥ − ≥ ≥ − ≥ (assumed a b c≥ ≥ )

The efficiency of EMV method has been shown in above examples – symmetric inequalities

with 3 variables. However, permutation inequalities with 3 variables are the one that EMV

mostly applied to:

Problem 10.6. Given , , 0a b c ≥ such that 3a b c+ + = . Prove that 2 2 2 4a b b c c a abc+ + + ≤

Proof

First of all, in order to use EMV, we need to homogenousize the inequality:

Given , , 0a b c ≥ .Prove that 2 2 2 34( )

27a b b c c a abc a b c+ + + ≤ + +

⇔ 38 6 3 ( ) 27 ( ) 0a abc bc b c bc c b− − + + − ≥� � �

( ) 2, , ( 4 4 )( ) 27( )( )( ) 0f a b c a b c b c a b b c c a⇔ = + + − + − − − ≥�

Thus: ( , , ) ( , , ),f a b c f a x b x c x≥ − − − { }0, min , ,x a b c� ∀ ∈�

So we only need to prove the inequality when c = 0: 2 34( )

27a b a b≤ +

Using AM – GM, we have:

3

2 342 24 4 ( )2 2 3 27

a a ba aa b b a b

� �+ +� �= ⋅ ⋅ ⋅ ≤ = +� ��

Therefore the problem is solved.

Equality occurs ⇔ 1a b c= = = or 2, 1, 0a b c= = = or its cyclic permutations

Note: While we use “mixing variables”, the expression ( )( )( )a b b c c a− − − appears to be

noticeable. Its role is to eliminate the permutative property of the rest of the inequality, which

make transforming to SOS form easier


181

Problem 10.7. Given distinct numbers , , 0a b c ≥ . Prove that:

2 2 2a b b c c a a b b c c ab c c a a b b c c a a b

− − − + + +� � � � � �+ + > + +� � � � � �− − − + + +� � � � � � (1)

Proof

Assume that { }min , ,c a b c= and let ( ) ( )( , , ) 1 1f a b c LHS RHS= − . We have

2( )( ) ( )

( )( ) ( )( )a b b c c a c a c b a bb c c a a b b c a b c a c b

+ + + − − −+ + = ++ + + + + + +

Therefore we have: ( , , ) ( , , ), [0, ]f a b c f a x b x c x x c≥ − − − ∀ ∈

We now only need to prove the inequality when 0c = :

2 2 2a b b a a b b a

b a a b b a a b− +� � � � � �+ + ≥ + +� � � � � �− +� � � � � �

Standardize 1b = , the inequality becomes ( ) ( ) ( )222 111 1

1 1aaa a

a a a a− + + > + + +

− +

This simple inequality is left for readers to solve.

Problem 10.8 [Le Trung Kien] Find all constants k such that

( ) 2 2 2 3 22

a b c a b c kka b b c c a ab ac bc

+ + −+ + ≤ ++ + + + +

, , , 0a b c∀ ≥

Proof

The inequality ⇔ ( ) ( ) ( )( ) ( ) ( )

( ) ( ) ( )2 2 2a b a c b c a b b c c akab ac bca b a c b c

− − − − + − + −≤+ ++ + +

( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( )

2 2 2a b a c b c a b a c b ck

ab ac bca b b c a c

− − − + + +⇔ ≤+ +− + − + −

(1)

Let ( ) ( ) ( ), , 1 1f a b c RHS LHS= − . We need to find k such that ( ), , 0 , , 0f a b c a b c≥ ∀ ≥

We will prove ( ) ( ) { }, , , , 0, min , ,f a b c f a x b x c x x a b c� ≥ − − − ∀ ∈� (2)

Indeed ( ) ( ), , , ,f a b c f a x b x c x≥ − − −

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

2 2 2a b a c b c a b x a c x b c xab bc ac a x b x b x c x c x a x

+ + + + − + − + −⇔ ≥+ + − − + − − + − −

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )3 a x b x c xabca b c a b c x

ab ac bc a x b x b x c x c x a x− − −⇔ + + − ≥ + + − −

+ + − − + − − + − −

( )21 13

1 1 11 1 1x

a b ca x b x c x

⇔ + ≥+ ++ +

− − −

(3)

Consider the function ( ) ( )3g x LHS= , then ( ) ( )0 3g RHS= . We have

( ) ( ) ( ) ( )

( )2 2

2

1 1 1

3 01 1 1

a x b x c xg x

a x b x c x

2+ +− − −′ = − >

+ +− − −

� ( ) ( )0g x g≥ . Thus (2) is true.


182

From (2) we have: ( ), , 0 , , , 0f a b c a b c≥ ∀ ≥ ( ), , 0 0 , , 0f a b a b⇔ ≥ ∀ ≥

Consider: ( ), , 0 0f a b ≥ ( )( )2 2 2

kab a b a ba b a b

−⇔ ≤ +− + +

(4)

We have (4) is true , 0a b∀ ≥ ( )mink h x⇔ ≤ where ( ) ( )( )

32 1 , 11

xh x xx x

+= ≥−

It is easy to see that ( ) ( )min 2 9 6 3 1h x x= + ≥ when 1 3 2 32

x + +=

Thus (1) is true , , 0a b c∀ ≥ ⇔ 2 9 6 3k ≤ +

Equality occurs a b c⇔ = =

Furthermore if 2 9 6 3k = ± + then the equality also occurs when ( ) ( ), , , ,a b c c b a� � is a cyclic

permutation of 1 3 2 3 , ,0 , 02

t t t� �+ +� � ≥� �

Remark: When we need to find optimal constants and when equality occurs correspondingly,

EMV is the perfect choice. All other methods either reach dead-end or are too complicated. In

the next section, we will see an interesting aspect of EMV.

2. EMV for inequalities with triangles.

We begin with the following problem:

Problem 10.9. Let , ,a b c be the length of edges in a triangle (possible to degrade)

Find the maximum value of ( , , )a b c

f a b cb c c a a b

= + ++ + +

Solution

The function ( , , )f a b c is associated with Nesbit inequality, and we know that

3min ( , , )2

f a b c = , when a b c= = .

However we need to find the maximal value and this problem is solved as following:

Assume { }max , ,a a b c= . After simple transformation:

( )2 3( , , )

2( )( ) 2a b c b c

f a b cb c c a a b a b a c

−= + + = ++ + + + +�

So ( , , ) ( , , ),f a b c f a x b x c x≤ − − − [0, ]x k∀ ∈ . Where k is a suitable number which will be found

later. So, we only need to prove the inequality where one variable is on the boundary. However,

is this problem’s boundary similar to previois problems? Let us investigate on this issue


183

Let try with x = k = min{a, b, c} i.e. one variable is 0, for example c = 0.

So ( , , 0)a b

f a bb a

= + . We cannot say anything about the maximal value in this case.

Note that c = 0 implies a b= because 0 | |c a b= ≥ − , so ( , , ) 2f a b c = .

This weakens ƒ(a, b, c) significantly, which make it impossible to find max ƒ(a, b, 0).

Therefore, we need to find better boundary. Note that, if max{ , , }a a b c= then the sufficient

and necessary condition for a, b, c to be length of edges in a triangles (possible to degrade) is

a b c≤ + . Thus b c+ is the boundary of a. We now begin to force a to b c+ .

We have 0a x b x c x x b c a− = − + − ⇔ = + − ≥

Let x k b c a= = + − , that is a b c= + . Therefore:

3( , , ) 1 2 2

2 2 (2 )(2 )b c bc

f b c b cc b b c b c c b

+ = + + = − ≤+ + + +

Equality occurs if and only if a = b, c = 0. So max ƒ(a, b, c) = 2

Problem 10.10. Given , , [2,3]a b c ∈ . Prove that:

( )1 1 1( ) 6a b c

a b ca b c b c c a a b

� �+ + + + ≥ + +� �+ + +� � (1)

Proof

To use EMV, we first need to change the given conditions to suitable. Note that for all

, , [2,3]x y z ∈ then ,3 2x y z x y+ ≥ ≥ . We will prove the inequality with the condition that

, ,a b c are length of edges of a triangle such that { } { }3min , , 2 max , ,a b c a b c≥ . We have,

(1) ⇔ ( ) ( )2 2 21 30 ( ) 2 ( ) 0

( )( )b c a b c a ab ac bc b c

bc a b a c� �− − ≥ ⇔ + + + − − ≥� �+ +� �

� �

( )2 2 2(3 )( )( ) 2 ( ) ( )( ) 0a a b c b c b c a b c a b a c b c⇔ − − + − − + − − − ≥� �

( ) ( ) ( )2 2 22 2(3 )( )( ) 2 0a a b c b c b c b c c a a b⇔ − − + − + − − − ≥� (2)

Let ( )( , , ) 2f a b c LHS= and assume max{ , , }a a b c= .

We will prove: ( , , ) ( , , ),f a b c f a x b x c x≥ − − − [ ]0,x b c a∀ ∈ + − .

This is obvious because 3 3( ) ( ) ( ),3 0a b c a x b x c x a b c− − ≥ − − − − − − − ≥ and similar inequalities.

So we only consider the case when a b c= + :

( ) ( ) ( ) 41 1 12( ) 6 1 02 2

b cb c b cb c b c c b b c

+ + + ≥ + + ⇔ − ≥+ + +

The inequality is proved. Equality occurs if and only if a b c= = .


184

Problem 10.11. Let a, b, c be length of edges of a triangle. Prove that

( ) ( )4 3 3a b c a c bb c a c b a

+ + ≥ + + + (1)

Proof

(1) ⇔ ( ) ( )7 6 0a b c a c b a c b a b cb c a c b a c b a b c a

+ + − − − + + + + + + − ≥

2 2 27( )( )( ) ( ) ( ) ( ) 0a b b c c a a b c b c a c a b⇔ − − − + − + − + − ≥

So we only need to prove when c a b= + , that is:

2 3 3 3 3 2 27 ( ) ( )( ) 0 2 2 7 7 0ab b a a b a b a b a b ab a b− + + − + + ≥ ⇔ + − + ≥

( )22 3772 2 0

4 8b b a a b a⇔ − + + ≥ . The inequality is thus solved.

Equality occurs if and only if a b c= =

Remark: Using EMV we can solved the generalized problem:

Let a, b, c be length of edges of a triangle. Find the best constant k such that:

( )1 3a b c b c a

k kb c a a b c� � � �+ + + ≥ + + +� � � ��

Problem 10.12 [Le Trung Kien]

Let a, b, c be length of sides of a triangle (possible to degrade).

Prove that: 5 5 5

9 9,5a b b c c ab c c a a b+ + +≤ + + ≤+ + +

(1)

Proof

5 5 59

a b b c c ab c c a a b+ + ++ + ≥+ + +

2( )( ) 10( )( )( )b c b c a b b c c a⇔ + − ≥ − − −�

It is easy to see that we only need to consider the case a b c= + , that is:

2 2 2( )( ) (2 ) ( 2 ) 10 ( )b c b c b c c b c b bc c b+ − + + + + ≥ −

( )23 2 2 3 2 3792 11 9 2 0 2 2 0

84b b c bc c c b c bc⇔ + − + ≥ ⇔ + + ≥−

Equality occurs when a b c= = .

On the other hand 25 5 5 ( ) 5( )( )( )

( , , )2( )( ) 2( )( )( )

a b b c c a b c a b b c c af a b c

b c c a a b a b a c a b b c c a+ + + − − − −+ + = + =+ + + + + + + +�

Now ( , , ) ( , , )f a b c f c b a≤ if a b c≤ ≤ . So we can assume that a b c≥ ≥ .


185

It is then obvious that ( , , ) ( , , ),f a b c f a x b x c x≤ − − − [ ]0,x b c a∀ ∈ + − .

So again, we only need to consider the case when a b c= + , that is:

6 5 6 5

9,5 ( 11 ) 02 2

b c b c c bbc b c

b c c b b c+ + ++ + ≤ ⇔ + ≥

+ + +

The problem is then solved. Equality occurs ⇔ , 0a b c= = and its permutation.

Note: It is interesting to see that 2-sided inequality can be solved completely using EMV

Problem 10.13. (Le Trung Kien) Let a, b, c be length of edges of a triangle

Prove that: 2 2 2 2 2 2

2a b c a b c

a b cb c a c a b

� �+ + ≥ + + + + +� �

� � (1)

Proof

(1) ⇔ ( )2 2 2 2 2 2 2 2 2 2 2 2

3 2 0a b c a b c a b c a b c

a b cb c a c a b b c a c a b

� �+ + − − − + + + + + + − + + ≥� �

� �

( ) ( ) ( ) ( )2 2 23( )( )( ) ( ) ( ) ( )0

a b b c c a a b c b c b c a c a c b a b aabc bc ac ba

− − − + + + − + − + −⇔ + + + ≥

( ) ( ) ( )2 2 2( ) ( ) ( )3( )( )( ) 0

a b c b c b a c a c c b a b aa b b c c a

a b c+ − + + − + + −⇔ − − − + ≥

+ + (2)

Assume max{ , , }c a b c= . We have: ( ) 1 1

1 1 1 12

a b ca b c

a b c a x b c x

+ = ≥+ + + +

+ − + −

, [ ]0,x a b c∀ ∈ + −

and similar inequalities, so we only need to prove (2) when c a b= + , that is:

( ) ( ) ( ) ( )2 23 32 2

3 ( ) 02( )

a b a b a b a b a bab b a

a b+ + + + + −− + ≥

+

( ) 24 3 2 2 3 4 2 2 32 4 0 2 0a a b a b ab b a ab b ab⇔ − − + + ≥ ⇔ − − + ≥

The problem is thus solved. Equality occurs ⇔ a b c= =

Remark: Using EMV we can solve the generalized problem:

Let a, b, c be length of edges of a triangle. Find the best constant k such that:

( )2 2 2 2 2 2

1a b c a b c

k k a b cb c a c a b

� � � �+ + + ≥ + + + + +� � � �

� � � �

Problem 10.14 [Le Trung Kien]

Let a, b, c be length of edges of a triangle (possible to degrade).

Prove that: ( ) ( )2

20 16a b c a b cab bc ca a b b c c a

+ + + ≤ + ++ + + + + (1)

Proof

We can see that equality in (1) does not occur when a b c= = . Our transformation has no

effect at all, therefore we need to come up with another way


186

(1) ⇔ ( ) ( ) ( ) ( )( ) ( ) ( )

2 2 2 2 8a b c ab bc ca a b b c a cab bc ca a b a c b c

+ + − + + − − −≤+ + + + +

As , ,a b c are length of sides of a triangle so 2 2 2 2 2 2 0a b c ab bc ca+ + − − − ≤ . Thus

If a b c≥ ≥ then the inequality is true.

If c b a≥ ≥ , then the inequality is equivalent to

( ) ( ) ( ) ( ) ( ) ( ) ( )2 2 22 2 2 8a b a c b c ab bc ca a b c a b b c c a

ab ac bc+ + + + + − − − ≥ − − −+ +

Consider 0 x a b c≤ ≤ + − and ( ) ( ) ( ) ( )2 2 2f x a b x b c x a c x= + − + − + − +

( ) ( ) ( )2 2 2

132 1 1 1

a b b c c a x a b c

a x b x c x

− + − + − � �+ − − − +� �+ +� �− − −� �

So ƒ(0) = LHS (1). We also have:

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( )

2 2 2 2

2 2 2 2 2 2

2

24 16 2 3 3 3

1 1 1

32 1 1 1

f x x a b c x a b c ab ac bc

a b b c c a a x b x c x

a x b x c x

′ = − + + + − + + + + +

� + + �− + − + − − − −+ − � �+ + �− − −�

( ) ( ) ( ) ( ) ( )2 2 22 2 2 2 424 16 2 3 3 33

x a b c x a b c ab ac bc a b b c c a� ≤ − + + + − + + + + + + − + − + −�

( ) ( ) ( )

( ) ( ) ( ) ( )

2

2 2 22 2 2

24 1642 3 3 33

a b c a b c a b c

a b c ab ac bc a b b c c a

≤ − + − + + + + −

� − + + + + + + − + − + −�

( ) ( ) ( ) ( ) ( )22 2 22 259 59 11 37 59 11 03 3

c c a b a b ab c b c a a b� � = − + + − + − = − − − − − ≤� �

Function ( )f x is decreasing for 0 x a b c≤ ≤ + − , therefore ( ) ( )0f f a b c≥ + − so we only

need to consider (1) when c a b= + , that is:

( ) ( ) ( ) 3 2 2

2 2

( ) 2 24 8 4 3 11 0

3a b a b b a

ab ab b a a a b aba b ab

+ + + ≥ − ⇔ + + ≥+ +

The problem is thus solved. Equality occurs ⇔ , 0a b c= = or its permutations.

Note: The power of EMV is fully illustrated in above problem. Note that the evaluation

( , , ) ( , , )f a b c f a x b x c x≥ − − − in previous problems are quite loose, while in this problem the

tightness is shown via the proof of ( ) 0f x′ ≤ .


187

XI. SOME SPECIAL MIXING VARIABLES TECHNIQUES.

One of the most attractive features of MV method is the diversity of techniques. This

can be illustrated via a number of problems that can be solved using special MV

techniques:

Problem 11.1. [Phan Thanh Nam] Given [ ], , 1,1x y z ∈ − and 0x y z+ + = .

Prove that: 2 2 21 1 1 3x y y z z x+ + + + + + + + ≥ (1)

Proof

Lemma: If 0; , , 1ab a b a b≥ + ≥ − then 1 1 1 1a b a b+ + + ≥ + + +

Proof: Indeed, square both sides of the inequality we have

( ) ( ) ( ) ( )2 2 1 1 2 2 1 1 1 1 0a b a b a b a b a b a b ab+ + + + + ≥ + + + + + ⇔ + + ≥ + + ⇔ ≥

Application: Among there numbers 2 2 2, ,x y y z z x+ + + there must be at least two numbers

with the same sign. WLOG, assume ( ) ( )2 2 0x y y z+ + ≥ . Using the lemma we have:

LHS (1) ( ) ( )2 22 2 2 2 2 21 1 1 1 1 1x y y z z x z z y z x≥ + + + + + + + + = + − + + + + +

( ) ( ) ( )2 222 2 21 1 1 1 1 1z z z x y z z z z≥ + − + + + + + = + − + + + +

Finally, it suffices to prove that ( )22 21 1 4z z z z− + + + + ≥

( )2 3 2 2 4 2 22 2 1 2 1 1 2 1 0z z z z z z z⇔ + − ≥ ⇔ − ≥ − + ⇔ − ≤ which is true because 1 1z− ≤ ≤ ,

The problem is then solved. Equality occurs ⇔ 0x y z= = = .

Problem 11.2. [Pham Kim Hung] Given , , , 0; 3a b c d a b c d≥ + + + = . Find the maximum

value of ( ) ( ) ( ) ( )2 3 2 3 2 3 2 3ab a b c bc b c d cd c d a da d a b+ + + + + + + + + + +

Solution

It is easy to see that ( ) ( ), , , , , , 0f a b c d f a b c d≤ + ( { }min , , ,d a b c d= )

The problem becomes: Given , , , 0; 3a b c d a b c d≥ + + + = . Find max:

( ) ( ) ( ) ( ) ( ) ( )2, , 2 3 2 2 2f a b c ab a b c bc b c a c b a c a c b= + + + + = + + + +

We have ( ) ( ) ( ) ( ) ( ) ( ), , , ,0 ; , , 0, ,f a b c f a c b bc a c b f a b c f b a c ab b a c− + = + − − + = − −

Therefore ( ) ( ) ( ){ }, , max , , 0 , 0, ,f a b c f a c b f b a c≤ + +

In addition: ( ) 2 2 2 2, , 0 2( ) ( ) 2(3 ) (3 ) ( )f a c b a c b a c b b b b b g b+ = + + + = − + − =

( ) 2 2 2 20, , ( ) 2( ) ( )(3 ) 2( ) (3 ) ( )f b a c a c b a c b a c a c a c a c g a c+ = + + + = + − − + + − − = +

where 3 2( ) 9 ( 3) ( 2 3) 6 3 6 3, 0g x x x x x x= − = − − + + ≤ ≥ .


188

Therefore: ( ) ( ) ( ){ }, , max , , 0 , 0, , 6 3f a b c f a c b f b a c≤ + + =

Equality occurs ⇔ 3 3, 3, 0a b c d= − = = = or its permutations

So ( )max , , 6 3f a b c =

Problem 10.3. Given 1, , , 22

x y z � ∈ ��

. Prove that 8 5 9y yx xz z

y z x x y z� � � �+ + ≥ + + +� � � ��

Proof

Let ( ), , 8 5 9y yx xz zf x y z

y z x x y z� � � �= + + ≥ + + −� � � ��

Assume that { }max , ,x x y z= . We have:

( ) ( ) ( ) ( )28 5

, , , , 8 2 5 2 0y xz x zy yx x z zf x y z f x xz z

y z z x y x xyz− −� � � �− = + − − + − = ≥� � � �

� � � �

( ) ( ), , , ,f x y z f x xz z� ≥ . Now let ,1 2xt tz

= ≤ ≤ we have:

( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )

22

2 222 2

2 2 2

1 2, , 8 2 3 5 2 3 8 2 3 5 3

1 8 6 5 1 4 5 28 51 2 1 1 2 0

x xz zf x xz z t tz x x z tt

t t t t t tt t t ttt t t

� � � �= + − − + − = + − − + −� � � ��

− + − − + −= − + − − + = = ≥

The problem is thus solved.

Equality occurs ⇔ x y z= = or ( ) ( )1, , 2,1,2

x y z = ; ( ) ( )1 11, , 2 ; , 2,12 2

XII. GENERAL MIXING VARIABLES THEOREM

In this section, we will introduce GMV theorem (General Mixing Variables) for n variables. This theorem almost contains all possibilities of “mixing variables”

1. We begin with some definition in n� .

Definition 1:

• The space n� is the set of ( )1 2, , ... nx x x x= where ,ix i∈ ∀� .

• A sequence ( ){ }1, 2, ,, , ...,m m m n mx x x x= in n� is defined to converge to ( )1 2, , ..., n

nz z z z= ∈�

if each sequence ,i mx converges to iz when , 1, 2,...,m i n→ ∞ ∀ = .

• Let nD ⊂ � . A function :f D → � is continuous on D if: for any sequence { }mx D⊂ and

z D∈ such that { }mx converges to z, then we have: ( )mf x converges to ƒ(z).

Definition 2: Let nD ⊂ � . We say:

• D is closed if for any sequence { }mx D⊂ and nz ∈� such that { }mx converges to z, then we

have z D∈ .

• D is bounded if there exists M such that: ( )1 2, , ... nx x x x D∀ = ∈ , then , 1, 2,...,ix M i n≤ ∀ = .


189

For example, a finite set is closed and bounded

The following theorem is fundamental to our results (the proof will be given at the end of this section)

Weierstrass theorem: Let D be a closed and bounded set in n� , and :f D → � is

continuous function. Then ƒ has global minima in D, i.e. there exists 0x D∈ such that:

( ) ( )0 ,f x f x x≤ ∀ ∈� . We also have similar result for global maxima.

� Comment: This theorem is a generalization of a familiar result: "Let [a, b] be a closed

interval in R and [ ]: ,f a b → � is continuous, then ƒ has global minima in [a, b]". Therefore,

Weierstrass theorem is intuitively understandable for us.

2. GMV Theorem: We assume that:

• D is a subset of n� , and Λ is a closed subset of D.

• :f D → � is an arbitrary continuous function such that f has a minima in Λ .

• DDTT k →:,...,1 are transformations such that Λ∈∀=== xxxTxT k ,)(...)(1 .

We will give some criteria such that the minima of f on Λ is also the global minima of ƒ on D.

Theorem GMV1: If

• { }1,

( ) min ( ( ) , \jj kf x f T x x D

=≥ ∀ ∈ Λ

• Dxkj ∈∀=∀ ,,1 we have lim ( )mjm

T x→∞

∈ Λ , where ( )0 1( ) , ( ) ( )m mj j j jT x x T x T T x−= =

then ( ) ( ){ }min ,y

f x f y x D∈Λ

≥ ∀ ∈ .

Moreover, equality does not occur on Λ\D if 1,

( ) min ( ( )), \jj kf x f T x x D

=> ∀ ∈ Λ .

Theorem GMV2: If

• D is closed and bounded in n� .

• 1,

( ) min ( ( )), \jj kf x f T x x D

=> ∀ ∈ Λ .

then ( ) ( ){ }min ,y

f x f y x D∈Λ

≥ ∀ ∈ , moreover, equality does not occur on Λ\D .

Theorem GMV3: If

• D is closed and bounded in n� .

• 1,

( ) min ( ( )), \jj kf x f T x x D

=≥ ∀ ∈ Λ

• There exists a continuous function jh : D → � such that: ( ) ( )( )j j jh x h T x> , Λ∈∀ \Dx .

then ( ) ( ){ }min ,y

f x f y x D∈Λ

≥ ∀ ∈ .


190

Proof

GMV1: Let’s consider arbitrary Dx ∈ . We have: 1,

( ) min ( ( ))jj kf x f T x

=≥ . By induction we

have1,

( ) min ( ( )),mjj k

f x f T x x D=

≥ ∀ ∈ . Since ƒ is continuous and Λ∈∞→

)(lim xTmm:

( ) ( )1, 1,

( ) lim min ( ( )) min lim ( ) minm mj jm m yj k j k

f x f T x f T x f y→∞ →∞ ∈Λ= =

� �≥ = ≥� ��

Furthermore, if 1,

( ) min ( ( ))jj kf x f T x

=> then we have

( ( )) min ( )j yf T x f y

∈Λ≥ implies ( )( ) min

yf x f y

∈Λ> .

GMV2: Using Weierstrass Theorem, there exists 0x D∈ such that ( ) ( )0f x f x≤ , Dx ∈∀ .

If 0x ∉ Λ then 0 01,( ) min ( ( ))jj k

f x f T x=

> , contradiction. Therefore 0x ∈ Λ .

GMV3: Consider 0y ∈Λ such that ( ) ( ){ }0 Miny

f y f y∈Λ

= . We assume that there exists z∈D

such that ( ) ( )0f z f y< . Since )(xh j is lower-bounded on D, by adding necessary constant,

we may assume that ( ) 0, , 1,...,jh x x D j k≥ ∀ ∈ ∀ = .

Select ε > 0 small enough we will have ( ) ( ) ( )01

k

jj

f z h z f y=

+ ε <� .

Let ( ) ( ) ( ){ }1,

min ,jj kg x f x h x x D

== + ε ∀ ∈ then :g D → � is continuous and

( ) ( )( )1,

min , \jj kg x g T x x D

=> ∀ ∈ Λ and ( ) ( ) ( ){ }0 Min

yg z f y g y

∈Λ< ≤ .

This is contradiction with Theorem 2.

� Comment: GMV theorems look simple however it has a wide range of applications. For

each transformation, we have a corresponding “mixing variables” technique. For instance, we

have following collaries

Consequence 1: (Pham Kim Hung, SMV-Strongly Mixing Variables) Let:

• nD ⊂ � is a closed and bounded set and ( )0 , , ...,s s s s D= ∈ .

• Transformation T is defined as: For each tuple ( )1 2, ,... na a a D∈ , we select the largest and

smallest and then replace them by their average number. We assume that T: D → D.

• :f D → � be a symmetric and continuous function such that: ( ) ( )( ) ,f a f T a a D≥ ∀ ∈ .

Then ( ) ( )0 ,f a f s a D≥ ∀ ∈

Proof: Let ( ) 21 2

1

, ,...,n

n ii

h a a a a=

=� , we have RDh →: is continuous and

0( ) ( ( )), \ { }h a h T a a D s> ∀ ∈ . Then the theorem is proved using GMV3.


191

� Remark:

We can also use GMV1, we only need to check for all Da ∈ then 0lim ( )m

mT a s

→∞= .

This is intuitively clear, so we leave the proof as a small exercise for readers. In fact, we can

overlook the assumption that D is closed and bounded (i.e. Consequence 1 is true for all

subset D of n� such that DDT ⊂)( ).

Consequence 2: (Dinh Ngoc An, UMV − Undefined Mixing Variables) Given:

• ( )1 21

, ,... | 0, onstn

nn i i

i

D x x x x x x c=

� �� ⊂ = ∈ ≥ =� ��

�� . Let Λ is a set of elements in D such that t

coordinates are 0 and the rest are equal ( )0t ≥ .

• :f D → � is continuous and symmetric such that:

( ) ( )1 2 1 21 2 3 1 2 3, ,..., min , , ,..., , 0, , ,...,

2 2n n n

a a a af a a a f a a f a a a a

� �+ +� �≥ +� ��

then ( ) ( ){ }min ,y

f x f y x D∈Λ

≥ ∀ ∈

Proof: Consider the transformations 1 2, :T T D D→ as following:

1 2( ) ( ) ,T a T a a a= = ∀ ∈Λ and for each ( )1 2, , ... \na a a a D= ∈ Λ , select two indices i j≠

such that { }min 0, 1,...,i ta a t n= > = and { }1 2max , ,...j na a a a= , corresponding to 1T we

replace ,i ja a by its average, corresponding to 2T we replace ,i ja a by ),0( ji aa + . Then

( ) ( ){ } DaaTfaTfaf ∈∀≥ ,)(,)(min)( 21 .

Let ( ) 21 1 2

1

, ,...n

n ii

h a a a a=

=� and ( )2 1 2 1 2, , ..., max{ , ,..., }n nh a a a a a a= − then RDhh →:, 21 are

continuous and }{\)),(()( 0sDaaThah jjj ∈∀> , 2,1=∀j . The problem is then solved using GMV3.

� Remark: We can also check if for all Da ∈ then 1lim ( ) ( , ,..., )m

mT a s s s

→∞= and

2lim ( ) (0,...,0, ,..., )m

mT a r r

→∞= and apply GMV1.

3. Following problems will illustrate the power of GMV.

Problem 12.1. (Cauchy inequality) Given n non-negative numbers 1 2, , ... nx x x .

Prove that: 1 2 1 2... ...nn nx x x n x x x+ + + ≥ ⋅

Proof By normalizing, we assume that 1 2 ... 1nx x x = thus we need to prove 1 2 ... nx x x n+ + + ≥ .

Obviously we only need to consider ,ix n i≤ ∀ .

Let ( ) [ ]{ }1 2 1 2, , ... | 0, , ... 1n i nD x x x x x n x x x= = ∈ = , D is closed and bounded

Let ( ){ }0 1,1,1,...,1xΛ = = . Consider a continuous function :f D → � such that:

For each ( )1 2, , ... nx x x x D= ∈ then ( )1 2 ... nf x x x x= + + + .


192

Let T: D\Λ → D be: For each ( )1 2, , ... \nx x x x D= ∈ Λ then there exists i jx x≠ and we T(x) is

the tuple x after replacing ix and jx by their geometric mean.

Then ( ) ( )( ) ( )20i jf x f T x x x− = − > . Using GMV2 we have ( ) ( )0 ,f x f x x D≥ ∀ ∈ ,

Equality occurs if 0x x= .

Problem 12.2. (Dinh Ngoc An) Given k > 0, 1 2, ,... 0nx x x ≥ and 1 2 ,... nx x x n+ + = . Prove that:

( )2 3 1 3 1 2 1( ... ) ( ... ) ... ( ... ) max ,1

kk k k

n n nnx x x x x x x x x n

n−� �

+ + + ≤ � �−� �

Proof

Define 1 2 2 3 1 3 1 2 1( , ,..., ) ( ... ) ( ... ) ... ( ... )k k kn n n nf x x x x x x x x x x x x −= + + + .

WLOG, we may suppose that 21 xx ≤ . Then tsxtsx +=−= 21 , where ],0[ st ∈ .

Consider 2 21 2( , ,..., ) ( ) ( ) ( ) ( )k k k

nf x x x g t A s t s t B s t� = = + + − + −�

where 3 4 5 3 5 3 4 1( ... ) , ( ... ) ( ... ) ... ( ... )k k k kn n n nA x x B x x x x x x x x x −= = + + + .

We have: 1 1 2 2 1( ) ( ) ( ) 2 ( ) 0k k kg t Ak s t s t Bkt s t− − −� ′ = + − − − − ≥�

1 1 2( ) ( ) ( ) 0k k Bt

h t s t s tA

− −⇔ = − − + − ≥

1 12( ) ( 1) ( ) ( ) , ( ) ( 1) ( ) ( ) 0k k k kBh t k s t s t h t k k s t s tA

− − − − − −� � ′ ′′= − − + + − = − − − + >� �

Hence h(t) is convex, moreover +∞==−→

)(lim),0(0 thhst

. Therefore, either h(t) is always

positive (0,s), or h(t) changes its sign from the negative sign to the positive one on (0,s).

But g’(t) has the same sign with h(t), so { }( ) max (0), ( )g t g g s≤ . It follows that:

1 2 1 21 2 3 1 2 3( , ,..., ) max , , ,..., , (0, , ,..., )

2 2n n n

x x x xf x x x f x x f x x x x

� �+ +� �≤ +� ��

Applying the UMV theorem, we have the proof.

� Remark: We observe that if there is an equality dependency among n variables, we can fix

(n – 2) variables and let other two vary. Therefore, consequence 3 allow us see how n

variables vary by looking at how two variables vary. If there is more equality dependency

among variables, the number of varied variables can be more to preserve the dependency. We

often see only inequality with one equality dependency. We will introduce two problems

where there is more than one such dependency.


193

Problem 12.3. (Phan Thanh Viet)

Let 3≥n , consider the set D consisting of all tuples of n positive numbers satisfying

Aaaa n =+++ ...21 , Baaa n =+++ 222

21 ... , where A, B are given constants such that 2AnB >

Prove that:

a) D has a unique element ),...,,( 21 nbbb satisfying nbbb === ...32 .

b) The expression 1 2 1 2( , ,..., ) ...n nf a a a a a a= attains its maximum value in D if 1 2( , ,..., )na a a is

a permutation of ),...,,( 21 nbbb .

Proof

a) Assume that ),...,,( 21 naaaa = satisfying naaa === ...32 . To have Da ∈ we need

0, 21 >aa , Aana =−+ 21 )1( , Bana =−+ 22

21 )1( . It follows that 2a is a positive root of

the equation ( ) 2 2 22 2 2 2( 1) ( 1) ( 1) 2 ( 1) 0A n a B n a n n a A n a B− − = − − ⇔ − − − − =

Since this equation has a unique positive root, we are done.

b)

� Step 1: We first consider the case n = 3.

The equality ( )3 3 3 2 2 2 21 2 3 1 2 3 1 2 3 1 2 3 1 2 3

13 3( ) ( )2

a a a a a a a a a a a a a a a� + + − = + + + + − + +� implies

that 1 2 3 1 2 31( , , ) 3 (3 )2

f a a a a a a A B A= + − , hence ),,( 321 aaaf attains its maximum value on D

if two of the three variables ia are equal (readers can prove this by investigating a function in

one variable – see the article on ABC method of Nguyen Anh Cuong).

� Step 2: We now consider the general case. Let Λ be the set of all tuples in D with (n – 1)

equal components. Consider the transformation DDT →Λ\: as follows. Assume that

Λ∈= \),...,,( 21 Daaaa n , we may always choose 3 numbers, say ),,( 321 aaa , such that they

are pairwise distinct. Then, there exists a unique triple of positive numbers ),,( 321 eee such

that 321321 aaaeee ++=++ , 23

22

21

23

22

21 aaaeee ++=++ . The transformation T replaces

),,( 321 aaa by ),,( 321 eee . Obviously, DaT ∈)( , so by the result in the case n=3 we have

)())(( afaTf > . It follows that )(af attains its maximum value if Λ∈a .

That is exactly what we want to prove.


194

Problem 12.4. (Phan Thanh Nam)

Let n ≥ 3, consider the set D consisting of all tuples of n positive real numbers whose sum is

A and whose product is B, where A, B are given constants such that BnA nn > . Prove that

a) D has precisely two elements ),...,,( 21 nbbb , ),...,,( 21 nccc satisfying

nbbbb ===> ...321 and ncccc ===< ...321 .

b) The expression 2 2 21 2 1 2( , ,..., ) ...n nf a a a a a a= + + + attains its maximum value (or minimum)

in D if ),...,,( 21 naaa is a permutation of ),...,,( 21 nbbb (or ),...,,( 21 nccc , respectively).

Proof

a) Assume that ),...,,( 21 naaaa = satisfying naaa === ...32 . To have Da ∈ we need

0, 21 >aa , Aana =−+ 21 )1( , Baa n =−121 . It follows that 2a is a positive root of the equation

12 2 21

2

( 1) ( 1) 0n nnBA n a n a Aa B

a−

−− − = ⇔ − − + = . Consider 1( ) ( 1) n nd x n x Ax B−= − − +

2( ) ( 1) ( )nd x n x nx A−′ = − − � d(x) decreases on ( )0; An

and increases on ( );An

∞ .

Moreover, ( ) ( )(0) 0n

A Ad B B dn n

= > > − + = , hence the polynomial d(x) has precisely two

positive roots, one is in ( )0; An

while the other is in ( );An

∞ . We then have the proof.

b)

� Step 1: We first consider the case n = 3. We show that for all Daaaa ∈= ),,( 321 , we will

have ],[ 11 bcai ∈ where i = 1, 2, 3. It suffices to consider i = 1.

We have : 2 22 3 2 3 1

1

44 ( ) ( )

Ba a a a A a

a= ≤ + = − BaAa 4)( 2

11 ≥−⇔

The polynomial BxAxxg 4)()( 2 −−= has 3 roots, where the two smaller roots are 1b and 1c

Indeed, since BbbAbb ==+ 22121 ,2

1

22

21

44)(

bB

bbA ==−� , 1b is a root of g(x) and

similarly so is 1c , moreover by Viete's theorem the sum of three roots is 1122 cbA +>

(otherwise 1122 cbA +≤ 211 422 bbAc =−≥� 16

121

221

21

221

2

bcbb

ccbb

c ≤≤=� , contradicts

1212 23 bAccc >=+≥ ) so the other root is greater than 1b .

Hence ],[0)( 1111 bcaag ∈⇔≥ . In particular, ],[, 1122 bccb ∈ 1221 bcbc <<<� .


195

We now have 2 2 2 2 2 2 21 2 3 1 2 3 2 3 1 1

1

2( ) 2 ( ) Ba a a a a a a a a A aa

+ + = + + − = + − − .

Consider xB

xAxxh2

)()( 22 −−+= where ],[ 11 bcx ∈ . We have: 2

2 )2(2)('

xBAxx

xh+−=

As in the remark, we see that the equation 0)(' =xh has exactly 2 positive roots, those are 2b

and 2c , moreover h’(x) is negative on ),( 22 cb and positive on ),(),( 1221 bcbc ∪

On the other hand, if { }211 ,bba ∈ then ),,( 321 aaa is a permutation of ),,( 321 bbb , hence

)(),,()( 23211 bhbbbfbh == , similarly )(),,()( 23211 chcccfch == . It then follows that

),,(),,(),,( 321321321 bbbfaaafcccf ≤≤ . Moreover, equality occurs if ),,( 321 aaa is a

permutation of ),,( 321 bbb or ),,( 321 ccc respectively (q.e.d).

� Step 2: We now consider the general case. We prove for the minimum case, the maximum

one is similar. Let Λ be the set of all permutations of ),...,,( 21 nccc , that is the set of all tuples

in D with (n – 1) equal components and greater than the remaining one. Consider the

transformation DDT →Λ\: as follows.

Assume that Λ∈= \),...,,( 21 Daaaa n , we may always choose 3 numbers , say ),,( 321 aaa ,

such that the case that two numbers are equal and greater than the remaining one doesn’t hold.

Then there exists a unique triple ),,( 321 eee such that 321 eee =< ,

321321 aaaeee ++=++ , 321321 aaaeee = . The transformation T replaces ),,( 321 aaa by

),,( 321 eee . Obviously DaT ∈)( . So by the result in the case n = 3, we have )())(( afaTf >

It follows that ƒ(a) attains its maximum value if Λ∈a .

That is exactly what we want to prove.

� To end this section, we give a proof for Weierstrass Theorem.

Definition 3: Given a sequence { } 1m ma ∞

= (in � or �n). A sequence { }

1km ka

∞

= is called a sub-

sequence of { } 1m ma ∞

= if { } 1k k

m ∞=

is a strictly increasing sequence.

For example: { }2 1m ma ∞

= is a subsequence of { } 1m m

a ∞=

.

Lemma 1: (Weierstrass) Any bounded sequence { }ma in � has a convergent subsequence

Proof: We know that “if a sequence is monotonous and bounded then it converges”, thus we

only need to show there exist a monotonous subsequence.


196

Consider { }| : m mT m m m a a+′′= ∈ ∃ > ≥�

If T is finite then { }ma is decreasing from some index. If T is infinite then we can extract an

increasing subsequence. In both cases, we can show there exists a monotonous subsequence.

Lemma 2: (Weierstrass) Any bounded sequence { }ma in n� has a convergent subsequence

Proof: Let ( ){ }1, ,, ...,m m n ma x x= be a bounded sequence in n� . Since { }1,mx is bounded in �,

there exists a subsequence { }11, kmx that converges. The sequence { }1

2, kmx is also bounded

in� so there exists a subsequence { }22, kmx that converges. By selecting "subsequence of

subsequence" continuously, we finally have a subsequence ( ){ }1, ,, ...,k k km m n ma x x= such

that 1,i n∀ = , { }, ki mx converges in �. Therefore { }kma converges in n� .

Lemma 3: (Completeness of �) Let A be a bounded set in �. There exists M ∈� such that: M A≤

(i.e. ,M a a A≤ ∀ ∈ ) and a sequence { }ka in A converges to M. We will denote this by infM A= .

Proof: We will prove that ∀ε > 0, ∃a∈A, a – ε ≤ A. Assume this is not correct. Consider arbitrary

1x A∈ , by induction, we can build a sequence { }mx in A such that 1 ,m mx x m ++ ≤ − ε ∀ ∈�

So ( )1 1 ,mx x m m +≤ − − ε ∀ ∈� this is contradictory to the fact that A is lower-bounded.

Thus, m +∀ ∈� , there exists ma A∈ such that 1ma A

m− ≤ . Since { }ma is bounded, there

exists a subsequence { }kma that converges to M in �. What is left to prove is that M ≤ A.

Indeed, consider arbitrary a∈A we have 1 ,km

k

a a km

+− ≤ ∀ ∈� , when k → ∞ then M ≤ a.

Proof of Weierstrass Theorem: Let A = ƒ(D). We will prove that A has a minimum value.

We will show that A has following property: if the sequence { }ma is contained in A and ma → α

then Aα∈ . Indeed, by definition we have mx D∈ such that ( )m mf x a= → α . Since { }mx is

bounded (contained in D), there exists a subsequence { }kmx that converges to c in n� . Since

D is closed, we have c D∈ . Since ( )mf x → α we have ( )kmf x → α . On the other hand, { }kmx c→

and ƒ is continuous we have ( ) ( )kmf x f c→ . This limit is unique, therefore ( )f c = α .

Now, A is lower-bounded (since it will lead to contradiction if α = −∞ ). Therefore, there

exists M = inf A. By the definition of inf and the property of A we have shown, we have M∈A.

So A has a minimum value which is M. The theorem is proved.


197

XIII. REVIEW

Dear reader, it is necessary to review our journey so far. As mentioned in VI, MV method is

known early when considering convex function together with a lot of nice results. Jensen

inequality can be seen as a criteria for mixing variables to centre globally. For details, see

Inequalities by Hardy − Polya − Littewood.

Our purpose is not just to describe a wide variety of “mixing variables” techniques but also to

let the readers grasp the idea of the method. Similar to many other methods, sometimes we

might not solve the problem directly by MV method; we need to transform it to a suitable

form. For example

Problem 13.1. (VMEO I − Phan Thanh Nam)

Find all constant k such that there exists 0kc > satisfies:

( ) ( ) ( ) ( )2 2 21 1 1 kkx y z c x y z+ + + ≥ + + , ∀x, y, z > 0

For each k, find the best constant kc .

Using limit, it is easy to see that k∈ [0, 2], however it much more complicated to find best.

You might try and see that it is difficult to use MV method here with variables x, y, z.

However by transforming tg , tg , tgx y z= α = β = γ where ( ), , 0;2πα β γ ∈ then it is surprised

that we can “force” the inequality to one variable by replacing , ,α β γ by their average. The

rest is just to investigate the behavior of an one-variable function or simply use generalized

AM – GM inequality.

MV techniques we have introduced hopefully give the overview about MV method. This

method is widely applicable, from 3-variable, 4-variable to n-variable inequalities. Generally

speaking, inequalities with 3 or 4 variables are still the first customers of MV method. For

many general problems with n variables, “mixing variables” is really difficult. Theorems such

as SMV, UMV could not solve those problems. Sometimes, MV techniques could not solve

even symmetric and homogeneous inequalities with 3 variables:

Problem 13.2. Given a, b, c ≥ 0. Prove that

( ) ( )22 33 3 3 6 36 0a b c abc a b c abc� + + − + + + − ≥�

Equality occurs when (a, b, c) = (t, 2t, 3t), t ≥ 0 (and its permutations).

Then all MV techniques we have covered could not solve this problem.

Another example is the famous inequality Vasile.


198

Problem 13.3. Given real numbers cba ,, . Prove that: ( ) ( )22 2 2 3 3 33a b c a b bc c a+ + ≥ + +

Beside the case a b c= = , equality also occurs when 2 2 24 2sin , sin , sin

7 7 7a b c

π π π= = =

From this query, it is nature to raise a question that: “Is there any MV technique for those

problems?” This topic will be discussed with readers in near future.

With the possibility that optima can be at the centre or on the boundary, it is difficult to

reduce the number of variables to 1. Thus we hope that there is a global “mixing variables”

method, similar to Jensen inequality. With this goal, we presented two beautiful SMV (strong

mixing variables) and UMV (undefined mixing variables). These two theorems can be

thought to be “twin”: SMV is “specialized” in inequalities whose optima are at the centre,

UMV allows us to combine both cases – optima are either at the centre or on the boundary.

For simplicity, these two theorems are only considered for symmetric functions.

We realize that it is not necessary to separate two cases, in fact we can merge the strengths of

these theorems (as shown in consequence 3 / VIII). However, GMV theorem is not just

simply the generalization of SMV and UMV; it opens a new horizon with numberless ways of

“mixing variables”. What is required here is just: if a tuple ( )1 2, , ... nx x x x= is not in Λ, it can

be replaced by another tuple (T(x)). In SMV ("classical"), the event of forcing two variables -

the biggest and the smallest to be the same, we can have the feeling that n variables will reach

the average value; in this case it is not the case anymore. However, we are still able to

achieve the result without extra conditions.

Proposed Problems

Problem 1. (Mathlinks) Given a, b, c > 0 and abc = 1. Prove that 61 1 1 5a b c a b c

+ + + ≥+ +

Problem 2. (MOSP 2001) Let a, b, c be positive numbers such that abc = 1.

Prove that ( ) ( ) ( ) ( )4 1a b b c c a a b c+ + + ≥ + + −

Problem 3. Given , , 0a b c ≥ . Prove that:

( ) ( ) ( ) ( ) ( ) ( )3 3 3 2 2 22 2 1 2a b c ab bc ca abc a b c bc b c ac a c ab a b+ + + + + + + ≥ + + + + + + + +

Problem 4. Given x, y, z ≥ 0 satisfy 2 2 2 3x y z+ + = . Prove that ( )7 12 9xy yz zx xyz+ + ≤ +

Problem 5. Given a, b, c > 0 and abc = 1. Prove that 2 2 2 3 1 1 16

2a b c a b c

a b c� �+ + + ≥ + + + + +� ��


199

Problem 6.(Pham Kim Hung)

Prove that 2 2 2

1 1 1 44 4 4 a b ca bc b ca c ab

+ + ≥+ ++ + +

, ∀ a, b, c ≥ 0

Problem 7. (Murray Klamkin) Given a, b, c ≥ 0 such that 2a b c+ + = . Prove that

( ) ( ) ( )2 2 2 2 2 2 3a ab b b bc c c ca a+ + + + + + ≤

Problem 8. (China 2005) Let a, b, c > 0 and 13

ab bc ca+ + = . Prove that

2 2 21 1 1 3

1 1 1a bc b ca c ab+ + ≤

− + − + − +

Problem 9. Let a, b, c∈ [p, q] where 0 < p ≤ q. Find the maximum value of

a b cb c c a a b

+ ++ + +

Problem 10 (A generalized of RMO 2000) Let a, b, c ≥ 0 and a + b + c = 3.

Find the lowest constant k > 0 such that the following inequality holds:

k k ka b c ab bc ca+ + ≥ + +

Problem 11. (Mathlinks) Let a, b, c ≥ 0 and ab + bc + ca = 1. Prove that:

( ) ( ) ( )

2 2 2 2 2 2

2 2 21 1 1 5

2a b b c c a

a b b c c a+ + ++ + ≥

+ + +

Problem 12.(Vasile Cirtoaje) Prove that for a, b, c > 0: 18 8 8

a b cb c c a a b

+ + ≥+ + +

Problem 13. (Phan Thanh Nam) Let x, y, z ≥ 0 such that 1a b c+ + = . Prove that

2 2 2 2x y y z z x+ + + + + ≥

Problem 14. Given , , 0a b c ≥ . Prove that: 2 2 2 2 2 24 4 4 4ab bc ca ac ab bc ab ac bc

b c b a a c+ + + + + ++ + ≥

+ + +

Problem 15. (Vasile Cirtoaje) Let x, y, z ≥ 0 such that 2 2 2 1a b c+ + = . Prove that

91 1 11 1 1 2ab bc ca

+ + ≤− − −

Problem 16. [Le Trung Kien] Let a, b, c be length of sides of a triangle.

Find the best constant k such that: 2 2 2a b c a b c a b c

k kb c a ab bc ca c a b

+ ++ + + ≥ + + ++ +

Problem 17. (Phan Thanh Viet) Let x, y, z∈ [−1, 1] and x + y + z = 0. Prove that

2 2 27 7 71 1 1 39 9 9

x y y z z x+ + + + + + + + ≥


200

Problem 18 [Le Trung Kien]

Given , , 0a b c ≥ . Prove that: 2 2 2

2 2 23( ) 4a b b c c aa b c

b c a c a b ab bc ca+ ++ + + ≥+ + + + +

Problem 19. Given , ,a b c ≥ 0. Prove that: 48 48 481 1 1 15a b cb c a c b a

+ + + + + ≥+ + +

Problem 20. (Pham Kim Hung) Given a, b, c ≥ 0 and a + b + c = 3. Prove that:

( ) ( ) ( )4 4 4 3 6 2a b c abc a b b c c a+ + − ≥ − − −

Problem 21. Given a, b, c ≥ 0, a + b + c = 2. Find the maximum value of

( ) ( ) ( )2 2 2 2 2 2a ab b b bc c c ca a− + − + − +

Problem 22. (Le Trung Kien)

Given , , 0a b c ≥ . Determine the best constant for the following inequality

2 2 2

2

32 3( )

ka b c a b cka b b c c a a b c

+ ++ + + ≥ ++ + + + +

Problem 23. Let , , 0a b c ≥ . Determine the best constant for the following inequality

3 3 3

2 2 2 2 2 2 331 ; 14

a b c kabc k ka b b c c a ab bc ca

+ + + ≥ + = −+ + + +

Problem 24. Given a, b, c ≥ 0 such that 4ab bc ca abc+ + + = .

Prove that: a b c ab bc ca+ + ≥ + +

Problem 25. Given a, b, c ≥ 0 such that 1

min{ , , }4

ab bc ca ≥ . Prove that:

( ) ( ) ( ) ( )2 2 2 23

1 1 1 3

1 1 1 1a b c abc+ + ≥

+ + + +

Problem 26. Given a, b, c ≥ 0 such that 4ab bc ca abc+ + + = .

Find the best constant k such that: 2 2 2 3 ( 1)( )a b c k k ab bc ca+ + + ≥ + + +

Problem 27. (Le Trung Kien) Given , , 0a b c ≥ . Prove that

( )22 2 2 2 2 1 (2 2)( )a b c abc ab bc ca+ + + + + ≥ + + +

Problem 28. Given , , 0a b c ≥ such that 1abc = Prove that

( ) ( )2 2 2 9 10a b c ab bc ca a b c+ + + + + ≥ + +

Problem 29. (Phan Thanh Viet) Prove in any triangle that

( ) ( ) ( )2 2 22 132a b cm m m p a b b c c a� + + ≤ + − + − + −�


201

Problem 30.(Jackgarfulkel) Let ABC be an acute triangle. Prove that:

a) ( )4sin sin sin 1 sin sin sin2 2 2 3 2 2 2

C CA B A B+ + ≥ +

b) ( )4cos cos cos 1 cos cos cos2 2 2 2 2 23

C CA B A B+ + ≥ +

Problem 31. (Jackgarfulkel) Let ABC be a triangle. Prove that:

( )2cos cos cos sin sin sin2 2 2 3

B C C AA B A B C− −− + + ≥ + +

Problem 32. (Phan Thanh Nam) Let ABC be a non-obtuse triangle.

a) Prove that: 2 2 2sin sin sin sin sin sin 5 2 sin sin sinsin sin sin 4B C C A A B A B C

A B C+ + ≥ + + +

b) Prove or disprove the following inequality:

( )sin sin sin sin sin sin 5 4 3 3 5 cos cos cossin sin sin 2B C C A A B A B C

A B C+ + ≥ + −

Problem 33. [Pham Kim Hung] Given , , ,a b c d and 2a b c d+ + + = . Prove that:

( ) ( )4 4 4 4 4a b c d abcd a b c d+ + + − ≥ − −

Problem 34. Let a, b, c, d be real numbers such that 2 2 2 2 1a b c d+ + + = .

Prove that: 1 1 1 1 1 1 81 1 1 1 1 1ab bc cd da db ca

+ + + + + ≤− − − − − −

Problem 35.(Phan Thanh Nam)

Let x, y, z, t be real numbers such that { }max , , , 1xy yz zt tx ≤ . Prove that:

( ) 22 2 2 21 1 1 1 16xy y yz z zt t tx x x y z t− + + − + + − + + − + ≥ + − + −

Problem 36. (Phan Thanh Nam)

Let [ ], , , 1;1x y z t ∈ − satisfying 0x y z t+ + + = . Prove that:

2 2 2 21 1 1 1 4x y y z z t t x+ + + + + + + + + + + ≥

Problem 37. Given , , , 0a b c d ≥ and 2 2 2 2 1a b c d+ + + = .Prove that:

( ) ( ) ( ) ( )1 1 1 1a b c d abcd− − − − ≥

Problem 38. Given , , , 0a b c d ≥ . Prove that:

3 3 3 3 34( ) 15( ) ( )a b c d abc bcd cda dab a b c d+ + + + + + + ≥ + + +

Problem 39. Given 1 2, , ..., 0na a a ≥ such that 1 2 ... 1na a a = . Prove that:

( )( ) ( )2 2 21 2 1 21 ... ( 3) (2 2) ...n nn x x x n n n x x x− + + + + + ≥ + + + +

Problem 40.Find the best constant km such that the following inequality holds

( ) ( ) ( ) ( ) ( )1 2 1 21 1 ... 1 1 ... 1nn m nmx mx mx m k x x x+ + + ≤ + + −

for every 1 2, , ... nx x x such that 1 2 ... nx x x n+ + + = and m is an arbitrary constant.


202

Problem 41. (Phan Thanh Viet)

Let 1 2, , ... , ,na a a s k > 0 such that: 1 2 ... nna a a s= and

( )1

1 knns

− =+

.

Consider the inequality: ( ) ( ) ( )1 2

1 1 1... 111 1k k k

n

naa a

+ + + ≤ −++ +

a) Prove that the above inequality doesn’t hold true in general.

b) Prove that the above inequality is true if k = 1.

c) Find all values of k (depending on n) such that the inequality holds true.

Problem 43. Given 1 2, , ..., 0na a a ≥ such that naaa n =+++ ...21 . Prove that

2222

21

21

2 )2()...)(1(41

...11 −++++−≥��

�

��

�+++ nnaaan

aaan n

n

Problem 43. Given 1 2, , ..., 0na a a ≥ such that 1 2 ... 1na a a+ + + = . Prove that:

1 2 2

1 2 2

1 1 1 2... 2

1 1 1 3

a a an

a a a− − −

+ + + ≤ − ++ + +


Consider the set D consisting of n positive real numbers naaa ,...,, 21 such that 22

22121 ...... nn aaaaaa +++=+++ . Find the maximum value of nn aaaaaaf ...),...,,( 2121 = .


Consider the set D consisting of n positive real numbers naaa ,...,, 21 such that

Aaaa n =+++ ...21 , Baaa n =+++ 222

21 ... , where A, B > 0 are given and nBA <2 .

a) Find the smallest value of321

1...

11aaa

+++ .

b) Find the smallest value of 222

21

1...

11

naaa+++ .

Problem 46. (Tran Ho Thanh Phu)


Aaaa n =+++ ...21 , Baaa n =+++ 222

21 ... , Caaa n =+++ 33

231 ... ,

where A, B are given positive constants such that the set D is nonempty.

Find the greatest value of the expression naaa ...21

Problem 47 (Phan Thanh Viet)


1 2

2 3 1

... 1naa an

a a a+ + + = + . Find the maximum and the minimum values of

21 22 2 21 2

( ... )

...n

n

a a a

a a a

+ + ++ + +


§24.1 NESBIT −−−− SHAPIRO INEQUALITY

I. BRIEF HISTORY ABOUT NESBIT −−−− SHAPIRO INEQUALITY

In 1905 an English mathematician named Nesbit introduced the inequality

32

a b cb c c a a b

+ + ≥+ + +

∀ a, b, c > 0

In this paper fifteen different ways to prove the above inequality will be given to enrich

readers’ thinking of inequality. Nesbit Inequality is not difficult; but it is not easy to extend

the inequality for n positive real numbers.

In 1954, the mathematician Shapiro proposed the general form of the inequality:

• Problem: Given 1 2, , ... na a a > 0. Determine the inequality is true or false:

11 21 2

2 3 3 4 1 1 2

( , , ..., ) ...2

n nn n

n

a aa a nf a a a

a a a a a a a a−= + + + + ≥

+ + + + (1)

This problem is so famous that in the year 1990, the mathematicians must say again about its

proof history at Oberwolfach mathematical seminar about inequality. This is not really a

surprise because the Shapiro inequality is not valid for all integer n. During 53 years a lot of

mathematicians tried to find the solution for proving the inequality (1) in specific and general

cases but it was not until 1958 that a pupil of Nesbit named Moocden found the solution for

4,5,6n = stopping a long time of failure.

In 1963, Diananda proved that:

If (1) is wrong for some odd 0n n= then it is also wrong for all 0n n≥ ; and if (1) is correct

for some even 0n n= then it is also correct for all 0n n≤ .

He either gave a counter example to claim that (1) is wrong for 27n = , means (1) is also

wrong for all odd 27n ≥ . After that Djokovic proved (1) is valid for 8n = and Bajsanski

proved that (1) is valid for 7n = .

In 1968, Nowosad proved that (1) is valid for 10n = so (1) is also valid for 9n = according

to Diananda theorem.

In 1971, Kristiansen proved that (1) is valid for 12n = ; (1) is also valid 11n = Daykin and

Malcoln proved that (1) is wrong for 25n = .

In 1976, Godunova and Leni again proved that (1) is valid for 12n = ; Bushell also gave a

different counter example to claim that (1) is wrong for 25n = . Bushell and Craven either

proved that (1) is valid for all odd n 23≤ .

Nesbit −−−− Shapiro inequality 204

In 1979, Troesch and Searcy propose a proof for a very impressive result is that (1) is wrong

for all even n 14≥ and using computer, he pointed out that (1) is true for all even n 12≤ and

odd n 23≤ (this is not a mathematical proof).

In 1985, Troesch published his mathematical proof (1) for 13n = in the Math Computing

journal; and in the year 1989 he continued his proof for 23n = . It shows that nearly 40 years

after Shapiro proposed his problem, Troesch was the person who finished a long history

period with the result: “Inequality (1) was true for all even n ≤≤≤≤ 12 and odd n ≤≤≤≤ 23. And for

all another n, (1) is”.

Another approach for this problem is that we will find the minimum for (1).

��

In 1958 Ranikin proved that: For ( )1 20

inf ( , ,..., )i

n naf n f a a a

>= then

8

1

( ) ( ) 1lim inf .7.10

2n n

f n f nn n

−

→∞ ≥λ = = < i.e. (1) is false for n big enough

In 1969, a Russian 11th grade pupil named Drinfeljd used advanced mathematical methods to introduce an impressive result as follows:

11 21 2

2 3 3 4 1 1 2

... 0.989133 , , ,... 02

n nn

n

a aa a na a a

a a a a a a a a−+ + + + > × ∀ >

+ + + +

Drinfeljd’s method was the creation ϕ(x) as the convex envelope of the functions

1 e xy −= ; 2 2

2e ex x

y =+

. Then 1

(0) 0.4945668...2

λ = ϕ =

He, probaly thanks to the result, received mathematical reward Fields in 1990.

In 4 - 1991, Kvant Magazine was successful in proving the two following results:

1. ( )11 2

2 3 3 4 1 1 2

... 2 1n n

n

a aa an

a a a a a a a a−+ + + + > −

+ + + + = 0.41421356n, ∀ai > 0

2. 11 2

2 3 3 4 1 1 2

5...12

n n

n

a aa an

a a a a a a a a−+ + + + >

+ + + + = 0.41666666n, ∀ai > 0

Hereunder we will consider the best estimation of Shapiro Inequality in primary mathematics.

3. Let be given 1 2, , ... 0na a a > for 3 ≤ n ∈ ∞. Prove that:

S = 11 2

2 3 3 4 1 1 2

... n n

n

a aa a

a a a a a a a a−+ + + +

+ + + + > 0.4577996n (1)


Proof

Consider 0 < α < 1; k∈N*. Let 1 1 2 2;n na a a a+ += = ; ( )1

k

kαβ =

− α. We have:

( )1 21

1 11 2 1 2 1 2

kn ni ii i i

i ii i i i i i

a aa a aS k k

a a a a a a= ++

= =+ + + + + +

� �β α ++ α= = + ⋅ − β� �

+ + +� ��

( ) ( ) ( )( )

( )1 11 11

1 11 2 1 1 2 1

1 2kkk kn n

i i i ii i i ikk

i ii i i i i i i i

a a a aa a a ak k n k k n

a a a a a a a a+ ++ ++

= =+ + + + + +

� �� β α + β α ++ α + α� �= + ⋅ − β ≥ + ⋅ ⋅ − β� � � �+ + +� � +� � � ��

We will prove: ( ) ( ) ( ) ( )11 1 1 3k kk k

i i i i i ia a a a a a ++ + ++ α α + ≥ α +

( ) ( ) ( ) 11 1k kk ki i it t t +⇔ + α α + ≥ α + with

10i

ii

at

a

+= ≥

Consider the function ( ) ( ) ( ) ( ) 11 1k kk kf t t t t += + α α + − α + with t ≥ 0.

We have: ( ) ( ) ( ) ( ) ( ) ( )[ ]1 1 .... 1k mm kf t t t k k k m−= + α α + − − − +

( ) ( ) ( ) ( )[ ] ( ) ( ) ( )[ ]1 11 .... 2 1 1 .... 2k m k mk km t k k k m t k k k m− − + −+ ⋅ α α + − − − − α + + − −

� ( ) ( ) ( ) ( )[ ] ( ) ( )[ ]2 10 1 .... 2 1 1m k m k m kf k k k m k m m k− − += − − − − + α + α − + α

≥ ( ) ( ) ( ) ( )( ) ( ) ( )1 1 2 11 .... 2 1 1 0k k m k m m k m kk k k m k k

+ − − + + − +� �− − + + α − + α =� �

Since ƒ(t) is a k-degree polynomial then ( ) ( ) ( ) ( )0 0k kf t const f= = ≥

� ( ) ( ) ( ) ( )1 1 0 0k kf t f− −≥ ≥ � ( ) ( ) ( ) ( )2 2 0 0k kf t f− −≥ ≥ , ... , ( ) ( )0 0f t f≥ =

We have ƒ(t) ≥ 0 ∀t ≥ 0 � ƒ(ti) ≥ 0 ∀ti ≥ 0 � (3) is true.

From (2) and (3) follow:

( ) ( ) ( )1 11 11

1 11 2 1 2

1 1n nk

k k i i i ik kk

i ii i i i

a a a aS k k n k k n

a a a a+ ++ ++

= =+ + + +

� �+ +≥ + ⋅ β α ⋅ − β = + αβ ⋅ − β� �

+ +� ��

( ) ( ) ( )( )[ ]

1

1

1 111

k kkk

kk

k k n k nk

+

+

� �α α� �≥ + αβ − β = + −� �� − α� �− α� �

Let 1 11

kx +=− α

� 1

11

kx +α = − , then ( ) ( )( )[ ] 1

111

k k

kk

g x kk +

α α= + ⋅ −− α− α

( ) ( ) ( )2

11

11

1 1 11 1

kk k kk k

k k kk

k xg x x x k k x

x xk

+ −+

++

� �+ − � �= − − = + −� ��

Nesbit −−−− Shapiro inequality 206

x 1 x0 + ∞

g′(x) 0

g(x) Max

( ) ( ) ( ) ( ) ( ) ( )2 2

2 2

11 2 1 1 11

2

1 1 . 1 11

k k kk k k k k kkk

k k

k x k x x k x x xg x k k x

x x

−+ − + +−+� �− + − − −′ = + − −� �

We have: ( ) ( ) ( ) ( )2 2

1 2 1 11

1

1 k 1 10 1 0

k k kkk

k k

k x k x xg x k k x

x x

+ + +−+

+

� �+ − − −′ = ⇔ + − − =� �

⇔ ( )[ ] ( ) ( )1 2 1 2 111 k 1 1 0k

k k kkk x k x k k k x x x−

+ + ++� �+ − + + − − − =� �

⇔ ( ) ( )1 211 0k

k kkk x k k k x x x−

+ ++� �+ + − − + =� � ⇔ ( ) ( )1

1 211k kkx k k k kx x x+ ++� �+ + − = −� �

⇔ ( ) ( )21

1 2 2 21 11 1k

k k kk kk k x k k kx k x x x+

+ + ++ ++ + + − − = −

⇔ ( ) ( ) ( ) ( )21

2 1 21 11 1 1 1 0k

k kk kk x k k x k x k k+

+ ++ ++ − + + − − + =

⇔ ( ) ( )21

2 11 11 0k

k kk kh x x k x k x k+

+ ++ += − + − − =

We have: ( ) ( ) ( ) ( )1

1 12 1 1k kkh x k x k k x k+ +′ = + − + + −

( ) ( ) ( ) ( ) ( ) ( ) ( )1 11111 1 112 21 1 1 1 1

kkkk k kk k kkk kk x k k k x k k k x

k k+++ + ++

� �+ +≥ + ⋅ ⋅ − − + = + − −� ��

We will prove: ( ) ( )111 112 1 0

kk kkk k k

k+ +++ − − ≥ ⇔ ( ) ( )

11 12 1

kk kk k k+ ++ − ≥

⇔ ( ) ( ) 12 1k kk k k ++ − ≥ ⇔ ( )21

k kkk k+ ≥

− ⇔ ( ) 121 1

1

k

k k+ ≥ +

−

We have: ( ) 2 121 1 3 1 , 21

k

k kk k k

+ > + ⋅ = > + ∀ ≥−

� h′(x) > 0 � h(x) is increasing. But h(x) is continuous; ( )limn

h x→+∞

= +∞ ;

( )21 1

1 1 11 0k

k k kh k k k k+

+ + += − − < − < then h(x) = 0 has unique root x0 > 1

� g′(x) = 0 has unique root x0 > 1 � Variation Table � Max g(x) = g(x0)

where x0 is a root of the inequation

( ) ( )21

2 11 11 0k

k kk kh x x k x k x k+

+ ++ += − + − − =

and ( ) ( ) ( )2

111

1kk k

kk

xg x k k x

x

+ −+� �−= + −� �

If k = 500 then x0 = 1.013294063 and g(x0) = 0.4577996

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